NCERT Solutions · Class 11 Physics Gravitation

The idea: a hemispherical shell is exactly half of a complete spherical shell, and the field inside a complete spherical shell is zero everywhere. So at any point of the flat rim plane the two halves must cancel — and that cancellation is only possible if each half's field there is purely perpendicular to the plane, i.e. straight down into the bowl.What I read off the figure. The bowl (blue) is the hemispherical shell; all of its mass lies on the curved surface, on or below the flat elliptical opening at the top. The centre of that flat circular opening is marked \(\displaystyle c\), and the point \(\displaystyle P\) is drawn on the same flat plane but off to the left of \(\displaystyle c\) — that is the "arbitrary point". The four arrows at \(\displaystyle P\) are: \(\displaystyle g\) pointing up and to the right at about \(\displaystyle 45^\circ\) (out of the bowl), \(\displaystyle f\) pointing horizontally to the right (along the plane, towards \(\displaystyle c\)), \(\displaystyle e\) pointing vertically straight down, and \(\displaystyle d\) pointing down and to the left at about \(\displaystyle 45^\circ\). At the centre, \(\displaystyle b\) is straight up, \(\displaystyle a\) is horizontal, and \(\displaystyle c\) is straight down.Step $\displaystyle 1$ — What "gravitational intensity" means. The gravitational intensity (field) at a point is the force per unit mass placed there, \[\vec{E} \;=\; \frac{\vec{F}}{m_0}, \qquad \text{units } \mathrm{N\,kg^{-1}} \;(=\mathrm{m\,s^{-2}}), \] where \(\displaystyle \vec{F}\) is the gravitational force on a test mass \(\displaystyle m_0\). Each mass element \(\displaystyle dm\) of the shell contributes \[d\vec{E} \;=\; -\,\frac{G\,dm}{r^{2}}\,\hat{r}, \] with \(\displaystyle G\) the universal gravitational constant, \(\displaystyle r\) the distance from \(\displaystyle dm\) to the field point, and \(\displaystyle \hat{r}\) the unit vector pointing from \(\displaystyle dm\) to the field point. The minus sign says gravity is always attractive: every arrow \(\displaystyle d\vec{E}\) points from the field point towards the mass. The total \(\displaystyle \vec{E}\) is the vector sum over the whole shell.Step $\displaystyle 2$ — Set up coordinates on the picture. Let the flat rim plane be the plane \(\displaystyle z=0\), with \(\displaystyle +z\) pointing vertically up out of the bowl, and let \(\displaystyle x,y\) lie in that plane. The bowl occupies \(\displaystyle z\le 0\). Both \(\displaystyle c\) and \(\displaystyle P\) sit at \(\displaystyle z=0\).Step $\displaystyle 3$ — Use the shell theorem. The shell theorem states: the gravitational field at any point inside a uniform complete spherical shell is zero. So mentally complete the bowl into a full sphere by adding its mirror-image upper hemisphere (same radius, same uniform mass density, sitting in \(\displaystyle z\ge 0\)). \(\displaystyle P\) is then a point inside a complete uniform spherical shell, so \[\vec{E}_{\text{lower}}(P) \;+\; \vec{E}_{\text{upper}}(P) \;=\; \vec{0}. \tag{1} \] Here \(\displaystyle \vec{E}_{\text{lower}}\) is the field of the real bowl in the figure — the thing we want.Step $\displaystyle 4$ — Reflect, and read off the components. The plane \(\displaystyle z=0\) is a mirror plane for the completed sphere: reflecting \(\displaystyle z \to -z\) turns the lower hemisphere into the upper one, and it leaves \(\displaystyle P\) itself fixed, because \(\displaystyle P\) lies on the mirror plane. A vector reflected in that plane keeps its in-plane components and flips its normal component. So if \[\vec{E}_{\text{lower}}(P) = (E_x,\; E_y,\; E_z), \qquad\text{then}\qquad \vec{E}_{\text{upper}}(P) = (E_x,\; E_y,\; -E_z). \] Substituting both into equation ($\displaystyle 1$): \[(E_x,\,E_y,\,E_z) + (E_x,\,E_y,\,-E_z) \;=\; (2E_x,\; 2E_y,\; 0) \;=\; (0,\,0,\,0). \] Therefore \[E_x = 0 \ \mathrm{N\,kg^{-1}}, \qquad E_y = 0 \ \mathrm{N\,kg^{-1}}. \] The horizontal component of the bowl's field at \(\displaystyle P\) is exactly zero. Notice that nothing in this argument used where \(\displaystyle P\) sits in the plane — it works for the centre \(\displaystyle c\) and for any off-centre point equally, which is precisely why the question says "arbitrary point".Step $\displaystyle 5$ — Which way along the vertical? Equation ($\displaystyle 1$) only killed the horizontal part; the sign of \(\displaystyle E_z\) comes from attraction. Every mass element of the bowl lies at \(\displaystyle z \le 0\), i.e. at or below \(\displaystyle P\)'s own level, so every contribution \(\displaystyle d\vec{E}\) at \(\displaystyle P\) points downward or (for the elements right on the rim) horizontally. None of them points upward. Hence \[E_z \;<\; 0, \] and the resultant is directed vertically downward, perpendicular to the flat face, into the bowl.Step $\displaystyle 6$ — Match that to an arrow at \(\displaystyle P\). The only arrow at \(\displaystyle P\) that is vertically downward is \(\displaystyle \mathbf{e}\). Checking the others against what we derived: This is consistent with the previous part: at the centre \(\displaystyle c\) the field is the straight-down arrow \(\displaystyle c\), and we have now shown the direction stays straight down at every point of the flat face — only the magnitude changes as \(\displaystyle P\) moves.No numerical value is being asked for, so there is no significant-figure count to report; the answer is a direction, and it is normal to the flat face, pointing into the bowl.Answer: (ii) e — the gravitational intensity at \(\displaystyle P\) is directed vertically downward, perpendicular to the flat circular face of the shell (arrow \(\displaystyle e\)).

The neutral point lies on the straight line joining the Earth and the Sun, and it must fall between them — that is the only place the two pulls can point in opposite directions and cancel. Anywhere else along that line, both forces pull the rocket the same way (toward whichever body is farther, in a sense that never cancels), so only the segment between Earth and Sun can hold a zero-force point.Let the distance from the Earth's centre to this point be \(\displaystyle r \). Since the Earth–Sun separation (orbital radius) is \[R = 1.5\times10^{11}\ \text{m}, \] the distance from the Sun's centre to the same point is \(\displaystyle (R-r) \).Using Newton's law of gravitation, \(\displaystyle F = \dfrac{GMm}{d^2} \), where \(\displaystyle G \) is the universal gravitational constant, \(\displaystyle M \) is the attracting mass, \(\displaystyle m \) is the rocket's mass, and \(\displaystyle d \) is the separation — the pull of the Earth on the rocket is \[F_E = \frac{GM_e m}{r^2}, \] and the pull of the Sun is \[F_S = \frac{GM_s m}{(R-r)^2}, \] with \(\displaystyle M_e = 6\times10^{24} \) kg and \(\displaystyle M_s = 2\times10^{30} \) kg.At the neutral point these two forces are equal in magnitude (and opposite in direction, along the Earth–Sun line): \[\frac{GM_e m}{r^2} = \frac{GM_s m}{(R-r)^2}. \]Both \(\displaystyle G \) and the rocket's mass \(\displaystyle m \) cancel — the neutral point does not depend on the rocket at all, only on the two big masses and their separation. Rearranging, \[\frac{(R-r)^2}{r^2} = \frac{M_s}{M_e} \quad\Rightarrow\quad \frac{R-r}{r} = \sqrt{\frac{M_s}{M_e}}. \] (The positive square root is the physical one, since both \(\displaystyle r \) and \(\displaystyle R-r \) are positive lengths — the point sits strictly between the two bodies.)Substituting the masses: \[\frac{M_s}{M_e} = \frac{2\times10^{30}}{6\times10^{24}} = 3.33\times10^{5}, \qquad \sqrt{\frac{M_s}{M_e}} = \sqrt{3.33\times10^{5}} = 577.35. \]So \[\frac{R-r}{r} = 577.35 \quad\Rightarrow\quad \frac{R}{r} = 1 + 577.35 = 578.35 \quad\Rightarrow\quad r = \frac{R}{578.35}. \]This is the step that surprises people: even though the Sun is about \(\displaystyle 3.3\times10^{5} \) times heavier than the Earth, the neutral point is not a tiny fraction of the way from the Sun — it sits at roughly \(\displaystyle \tfrac{1}{578} \) of the Earth–Sun distance from Earth. That is because it is the square root of the mass ratio that sets the distance ratio, not the mass ratio itself; the inverse-square law tempers the huge mass difference into a much smaller distance factor.Putting in the number: \[r = \frac{1.5\times10^{11}\ \text{m}}{578.35} = 2.59\times10^{8}\ \text{m}. \]The input data (masses to one significant figure, orbital radius \(\displaystyle R = 1.5\times10^{11} \) m to two significant figures) supports at most two significant figures in the final answer, so this rounds to \[r \approx 2.6\times10^{8}\ \text{m}. \]This point lies on the Earth–Sun line, on the side of the Earth facing the Sun, at about \(\displaystyle 2.6\times10^{8} \) m from the Earth's centre (which is only about $\displaystyle 0.17$% of the total Earth–Sun distance — well inside the Moon's orbit, whose radius is about \(\displaystyle 3.8\times10^{8} \) m).Answer: The gravitational force on the rocket is zero at about \(\displaystyle 2.6\times10^{8}\ \text{m} \) from the Earth's centre, along the line joining the Earth and the Sun, on the side toward the Sun.

The Sun's gravity is exactly what keeps the Earth in its orbit, so equating that gravitational pull to the centripetal force needed for the orbit lets you solve for the Sun's mass — without ever standing on a scale.Newton's law of gravitation gives the force the Sun (mass \(\displaystyle M\)) exerts on the Earth (mass \(\displaystyle m\)), separated by the orbital radius \(\displaystyle R\):\[F_{\text{grav}} = \frac{GMm}{R^{2}} \]This is the only force holding the Earth in its (nearly circular) orbit, so it must equal the centripetal force required to keep the Earth moving on that circle with period \(\displaystyle T\) (one year):\[F_{\text{centripetal}} = m\omega^{2}R = m\left(\frac{2\pi}{T}\right)^{2}R \]Equate the two forces. Since it is the Sun's gravity that supplies the centripetal force, not some separate agent, set them equal:\[\frac{GMm}{R^{2}} = \frac{4\pi^{2}mR}{T^{2}} \]The Earth's mass \(\displaystyle m\) cancels from both sides — this is the whole point of the method: you can find the Sun's mass without ever needing to know the Earth's. Solving for \(\displaystyle M\):\[M = \frac{4\pi^{2}R^{3}}{GT^{2}} \]Substitute the numbers, converting everything to SI units first.Orbital radius: \(\displaystyle R = 1.5\times10^{8}\text{ km} = 1.5\times10^{11}\text{ m}\) (km → m needs a factor of $\displaystyle 1000$ — leaving it in km would make \(\displaystyle G\), which is defined in SI units, unusable).Orbital period: \(\displaystyle T = 1\text{ year} = 365.25 \times 24 \times 3600\text{ s} = 3.156\times10^{7}\text{ s}\).Gravitational constant: \(\displaystyle G = 6.67\times10^{-11}\ \text{N m}^{2}\,\text{kg}^{-2}\).First, cube the radius:\[R^{3} = (1.5\times10^{11}\text{ m})^{3} = 3.375\times10^{33}\text{ m}^{3} \]Numerator:\[4\pi^{2}R^{3} = 4(9.8696)(3.375\times10^{33}) = 1.332\times10^{35}\ \text{m}^3 \]Denominator — square the period first:\[T^{2} = (3.156\times10^{7}\text{ s})^{2} = 9.960\times10^{14}\text{ s}^{2} \] \[GT^{2} = (6.67\times10^{-11})(9.960\times10^{14}) = 6.643\times10^{4}\ \text{N m}^2\text{kg}^{-1}\text{s}^2 \]Now divide:\[M = \frac{1.332\times10^{35}}{6.643\times10^{4}}\ \text{kg} = 2.006\times10^{30}\ \text{kg} \]Significant figures. The orbital radius was given as \(\displaystyle 1.5\times10^{8}\) km — only $\displaystyle 2$ significant figures — so the final mass cannot be stated to more precision than that, even though the intermediate arithmetic carried more digits. Rounding to $\displaystyle 2$ significant figures:\[M \approx 2.0\times10^{30}\ \text{kg} \]This is roughly \(\displaystyle 3.3\times10^{5}\) times the Earth's mass, consistent with the Sun being the dominant mass of the solar system.Answer: The Sun's mass is \(\displaystyle M \approx 2.0\times10^{30}\) kg, found from \(\displaystyle M = \dfrac{4\pi^{2}R^{3}}{GT^{2}}\) using the Earth's orbital radius \(\displaystyle R = 1.5\times10^{11}\) m and period \(\displaystyle T = 1\) year, by equating the Sun's gravitational pull on the Earth to the centripetal force of the Earth's orbit.

Kepler's third law connects a planet's orbital period to its distance from the Sun — no need to know the mass of anything.Kepler's law of periods states that the square of a planet's orbital period \(\displaystyle T \) is proportional to the cube of its mean distance \(\displaystyle r \) from the Sun: \[T^2 = k\,r^3 \] where \(\displaystyle k \) is the same constant for every planet orbiting the Sun. Because \(\displaystyle k \) is common to both planets, it cancels when you compare Earth and Saturn — you never need its actual value.Set up the ratio.For Earth: \(\displaystyle T_e^2 = k\,r_e^3 \) For Saturn: \(\displaystyle T_s^2 = k\,r_s^3 \)Dividing the second equation by the first eliminates \(\displaystyle k \): \[\left(\frac{T_s}{T_e}\right)^2 = \left(\frac{r_s}{r_e}\right)^3 \]Solving for the distance ratio: \[\frac{r_s}{r_e} = \left(\frac{T_s}{T_e}\right)^{2/3} \]Substitute the numbers.You are told the Saturn year is $\displaystyle 29.5$ times the Earth year, so \(\displaystyle \dfrac{T_s}{T_e} = 29.5 \), and \(\displaystyle r_e = 1.50\times10^{8}\ \text{km} \).\[r_s = r_e \left(\frac{T_s}{T_e}\right)^{2/3} = \left(1.50\times10^{8}\ \text{km}\right)\times (29.5)^{2/3} \]Evaluate \(\displaystyle (29.5)^{2/3} \) by first cubing: \(\displaystyle (29.5)^2 = 870.25 \), then taking the cube root: \[(870.25)^{1/3} \approx 9.55 \]So: \[r_s \approx \left(1.50\times10^{8}\ \text{km}\right)\times 9.55 \approx 14.3\times10^{8}\ \text{km} \]Watch the exponent when you tidy up the power of ten. \(\displaystyle 14.3\times10^{8}\ \text{km} \) is the same number as \(\displaystyle 1.43\times10^{9}\ \text{km} \) — shifting the decimal point one place right in the coefficient means dropping the power of ten by one.Both given quantities — $\displaystyle 29.5$ and $\displaystyle 1.50$×$\displaystyle 10$⁸ km — carry three significant figures, so the answer is rounded once, at the end, to three significant figures: \[r_s \approx 1.43\times10^{9}\ \text{km} \]This is a distance of roughly \(\displaystyle 1.43\times10^{9}\ \text{km} \), or about $\displaystyle 9.5$ times the Earth–Sun distance, which is indeed close to Saturn's actual orbital radius — the huge period ratio (nearly $\displaystyle 30$ Earth years to one Saturn year) comes from a comparatively modest increase in distance, exactly as the \(\displaystyle 3/2 \) power relationship in Kepler's law predicts.Answer: The distance of Saturn from the Sun is \(\displaystyle r_s \approx 1.43\times10^{9}\ \text{km} \) (about $\displaystyle 9.5$ times the Earth–Sun distance).

Weight is just the gravitational force \(\displaystyle mg \), and that force falls off as the inverse square of distance from Earth's center — not as the inverse square of height above the surface. The height has to be added to the radius first.Step $\displaystyle 1$: Find the mass from the given weight.At the surface, weight is \[W = mg \] where \(\displaystyle g \) is the acceleration due to gravity at the surface. Given \(\displaystyle W = 63\ \text{N} \) and taking \(\displaystyle g = 9.8\ \text{m/s}^2 \), \[m = \frac{W}{g} = \frac{63\ \text{N}}{9.8\ \text{m/s}^2} = 6.43\ \text{kg} \]Step $\displaystyle 2$: Write the force law at a height above the surface.By Newton's law of gravitation, the force on a body of mass \(\displaystyle m \) at distance \(\displaystyle r \) from Earth's center (mass \(\displaystyle M \), radius \(\displaystyle R \)) is \[F = \frac{GMm}{r^2} \] At the surface (\(\displaystyle r = R \)) this reduces to \(\displaystyle F = mg \), since \(\displaystyle g = \dfrac{GM}{R^2} \). At height \(\displaystyle h \) above the surface, the distance from the center becomes \(\displaystyle r = R + h \), so \[F(h) = \frac{GMm}{(R+h)^2} = mg\left(\frac{R}{R+h}\right)^2 \] This is the step most people skip: they square the height instead of squaring \(\displaystyle (R+h) \), the actual distance from Earth's center.Step $\displaystyle 3$: Substitute \(\displaystyle h = R/2 \).\[R + h = R + \frac{R}{2} = \frac{3R}{2} \] \[\left(\frac{R}{R+h}\right)^2 = \left(\frac{R}{\dfrac{3R}{2}}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \]Step $\displaystyle 4$: Compute the force.\[F(h) = mg \times \frac{4}{9} = W \times \frac{4}{9} = 63\ \text{N} \times \frac{4}{9} \] \[F(h) = \frac{252}{9}\ \text{N} = 28\ \text{N} \]The mass itself never had to be pulled out explicitly — since \(\displaystyle W = mg \), the ratio \(\displaystyle F(h)/W = (R/(R+h))^2 \) applies directly to the given weight. Both $\displaystyle 63$ and the factor \(\displaystyle 4/9 \) carry two significant figures, so the result is properly stated to two significant figures.Answer: $\displaystyle 28$ N, directed toward the center of the Earth.

Inside a uniform sphere, only the mass closer to the center than you than pulls you — the shell theorem strips away everything farther out. For a point mass or a point outside a spherical shell, the shell theorem says a uniform spherical shell exerts zero gravitational force on anything inside it, and pulls on anything outside it exactly as if all its mass were concentrated at the center. So for a point at distance \(\displaystyle r \) from Earth's center (with \(\displaystyle r < R \), \(\displaystyle R \) = Earth's radius), only the mass \(\displaystyle M_r \) contained within the sphere of radius \(\displaystyle r \) contributes to the pull; the mass in the shell between \(\displaystyle r \) and \(\displaystyle R \) contributes nothing.Since the density \(\displaystyle \rho \) is uniform, \[M_r = \rho \cdot \frac{4}{3}\pi r^3, \qquad M = \rho \cdot \frac{4}{3}\pi R^3 \] so \[\frac{M_r}{M} = \left(\frac{r}{R}\right)^3 \quad\Rightarrow\quad M_r = M\left(\frac{r}{R}\right)^3. \]The acceleration due to gravity at that depth, using Newton's law of gravitation \(\displaystyle g(r) = \dfrac{GM_r}{r^2} \) (with \(\displaystyle G \) the universal gravitational constant), is\[g(r) = \frac{G}{r^2}\cdot M\left(\frac{r}{R}\right)^3 = \frac{GM}{R^2}\cdot\frac{r}{R} = g\left(\frac{r}{R}\right), \]where \(\displaystyle g = GM/R^2 \) is the acceleration due to gravity at the surface (\(\displaystyle r = R \)). This is the key result: inside a uniform Earth, \(\displaystyle g \) does not stay constant or follow the inverse-square law — it falls off linearly with distance from the center, reaching zero exactly at the center. (This is different from the region above the surface, where \(\displaystyle g \) falls as \(\displaystyle 1/r^2\); people often wrongly apply the surface formula all the way to the center.)"Half way down to the centre" means the point is at depth \(\displaystyle R/2 \) below the surface, i.e. at distance from the center\[r = R - \frac{R}{2} = \frac{R}{2}. \]Substituting into \(\displaystyle g(r) = g(r/R) \):\[g\!\left(\frac{R}{2}\right) = g\cdot\frac{R/2}{R} = \frac{g}{2}. \]Weight is just mass times the local value of \(\displaystyle g \), and the body's mass \(\displaystyle m \) does not change as it goes underground — only \(\displaystyle g \) changes. On the surface, \[W_{\text{surface}} = mg = 250\ \text{N}. \] Halfway down, \[W = mg\!\left(\frac{R}{2}\right) = m\cdot\frac{g}{2} = \frac{mg}{2} = \frac{W_{\text{surface}}}{2} = \frac{250\ \text{N}}{2} = 125\ \text{N}. \]The result has the same number of significant figures as the given weight ($\displaystyle 250$ N, three significant figures), so no rounding is needed beyond stating $\displaystyle 125$ N.Answer: $\displaystyle 125$ N (three-quarters of the way to zero remains to fall; the body weighs exactly half of its surface weight, since \(\displaystyle g \) inside a uniform Earth is proportional to the distance from the center).

The gravitational field falls off with height, so a rocket climbing to over a thousand kilometres cannot be treated with the constant-\(\displaystyle g\) formula \(\displaystyle v^2 = 2gh\) — you need conservation of energy with the full inverse-square potential.Since the rocket is fired straight up, its motion is purely radial. It will keep climbing, slowing down under Earth's pull, until its speed momentarily drops to zero — that instant marks the greatest height \(\displaystyle h\) it reaches, after which it falls back.Step $\displaystyle 1$: Check whether the rocket even returns.The escape speed from Earth's surface is \[v_{esc}=\sqrt{\dfrac{2GM}{R}} \] where \(\displaystyle G=6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2}\) is the universal gravitational constant (not to be confused with the surface acceleration \(\displaystyle g\), which is a derived quantity, \(\displaystyle g=GM/R^2\), and isn't fixed once you leave the surface), \(\displaystyle M=6.0\times10^{24}\ \text{kg}\) is Earth's mass, and \(\displaystyle R=6.4\times10^{6}\ \text{m}\) is Earth's mean radius.\[GM = (6.67\times10^{-11})(6.0\times10^{24}) = 4.002\times10^{14}\ \text{m}^3\text{s}^{-2} \] \[v_{esc}=\sqrt{\dfrac{2(4.002\times10^{14})}{6.4\times10^{6}}}=\sqrt{1.251\times10^{8}}\approx1.12\times10^{4}\ \text{m s}^{-1}=11.2\ \text{km s}^{-1} \]The rocket is launched at \(\displaystyle 5\ \text{km s}^{-1}\), well below \(\displaystyle 11.2\ \text{km s}^{-1}\), so it does not escape — it rises to a finite height and returns, which is exactly what the question is asking us to find.Step $\displaystyle 2$: Apply the law of conservation of mechanical energy.For a rocket of mass \(\displaystyle m\) moving only under gravity, the sum of kinetic energy and gravitational potential energy is conserved between the launch point (surface, distance \(\displaystyle R\) from Earth's centre, speed \(\displaystyle v\)) and the highest point (distance \(\displaystyle R+h\) from Earth's centre, speed zero): \[\frac{1}{2}mv^{2}-\frac{GMm}{R}=0-\frac{GMm}{R+h} \] The mass \(\displaystyle m\) of the rocket cancels — how far it goes does not depend on its mass, only on its launch speed. Dividing through by \(\displaystyle m\): \[\frac{1}{2}v^{2}-\frac{GM}{R}=-\frac{GM}{R+h} \] so \[\frac{GM}{R+h}=\frac{GM}{R}-\frac{1}{2}v^{2} \]Step $\displaystyle 3$: Substitute the numbers.With \(\displaystyle v=5\ \text{km s}^{-1}=5.0\times10^{3}\ \text{m s}^{-1}\): \[\frac{GM}{R}=\frac{4.002\times10^{14}}{6.4\times10^{6}}=6.253\times10^{7}\ \text{m}^2\text{s}^{-2} \] \[\frac{1}{2}v^{2}=\frac{1}{2}(5.0\times10^{3})^{2}=1.25\times10^{7}\ \text{m}^2\text{s}^{-2} \] \[\frac{GM}{R+h}=6.253\times10^{7}-1.25\times10^{7}=5.003\times10^{7}\ \text{m}^2\text{s}^{-2} \]Step $\displaystyle 4$: Solve for the distance, then the height.\[R+h=\frac{GM}{5.003\times10^{7}}=\frac{4.002\times10^{14}}{5.003\times10^{7}}\approx7.999\times10^{6}\ \text{m} \]This is the rocket's farthest distance from Earth's centre. To get the height above the surface — the distance the question is really asking about — subtract \(\displaystyle R\): \[h=(R+h)-R=7.999\times10^{6}\ \text{m}-6.4\times10^{6}\ \text{m}=1.599\times10^{6}\ \text{m} \]Every input in this problem carries two significant figures (\(\displaystyle 6.0\times10^{24}\), \(\displaystyle 6.4\times10^{6}\), \(\displaystyle 5.0\times10^{3}\)), so the answer should be rounded to two significant figures only at this last step, not earlier: \[h\approx1.6\times10^{6}\ \text{m} \]That is about $\displaystyle 1600$ km above the surface — roughly a quarter of Earth's own radius, which is why the constant-\(\displaystyle g\) shortcut would have given a noticeably wrong number here: the pull of gravity has weakened appreciably by the time the rocket gets that high.**Answer: The rocket rises to a height of about \(\displaystyle 1.6\times10^{6}\ \text{m}\) (≈$\displaystyle 1600$ km) above Earth's surface — equivalently, about \(\displaystyle 8.0\times10^{6}\ \text{m}\) from Earth's centre — before falling back.

Use energy conservation, not the escape-speed formula a second time. Once the body leaves the surface, gravity is the only force doing work on it, so its total mechanical energy — kinetic plus gravitational potential — stays constant all the way out to "far away" (where the potential energy is taken as zero).Let \(\displaystyle M \) be the earth's mass, \(\displaystyle R \) its radius, \(\displaystyle m \) the body's mass, and \(\displaystyle v_e \) the escape speed from the surface, \(\displaystyle v_e = 11.2 \text{ km s}^{-1} \). By definition of escape speed, \[\frac{1}{2}mv_e^{2} = \frac{GMm}{R} \] (a body launched at exactly \(\displaystyle v_e \) has just enough kinetic energy to cancel the gravitational potential energy at the surface, reaching infinity with zero speed left over).The body here is launched with three times the escape speed, \(\displaystyle v = 3v_e \). Energy conservation between the surface and a point far from the earth (where potential energy \(\displaystyle \to 0 \)) gives \[\frac{1}{2}mv^{2} - \frac{GMm}{R} = \frac{1}{2}mv_f^{2} + 0 \] where \(\displaystyle v_f \) is the speed far away. Divide through by \(\displaystyle m \) and use \(\displaystyle \dfrac{GM}{R} = \dfrac{1}{2}v_e^{2}\) from the escape-speed relation above: \[\frac{1}{2}v_f^{2} = \frac{1}{2}v^{2} - \frac{1}{2}v_e^{2} \] \[v_f^{2} = v^{2} - v_e^{2} = (3v_e)^{2} - v_e^{2} = 9v_e^{2} - v_e^{2} = 8v_e^{2} \] \[v_f = v_e\sqrt{8} = 2\sqrt{2}\,v_e \]This is the point people get wrong: the leftover speed is not \(\displaystyle \sqrt{9-1}=\sqrt8 \) times something arbitrary — it comes directly from subtracting the escape kinetic energy (which is fixed by \(\displaystyle R\) and \(\displaystyle M\)), not from simply scaling the launch speed.Substituting \(\displaystyle v_e = 11.2 \text{ km s}^{-1} \): \[v_f = \sqrt{8}\times 11.2 \text{ km s}^{-1} = 2.828 \times 11.2 \text{ km s}^{-1} = 31.678\ldots \text{ km s}^{-1} \]The given data (\(\displaystyle 11.2\), "thrice") carry three significant figures, so the result is rounded to three significant figures: \[v_f \approx 31.7 \text{ km s}^{-1} \]This is a speed, not a vector direction to report separately — the body moves radially outward along the same line it was launched on, simply retaining this much speed once it is far from the earth (with the sun and other planets ignored, as stated).Answer: The body still has a speed of about \(\displaystyle 31.7 \text{ km s}^{-1}\) when far away from the earth (moving radially outward, obtained from \(\displaystyle v_f = \sqrt{(3v_e)^2 - v_e^2} = 2\sqrt2\,v_e\)).

A bound satellite has negative total energy — sending it "out of Earth's influence" means supplying exactly enough energy to lift that total up to zero.For a satellite of mass \(\displaystyle m \) moving in a circular orbit of radius \(\displaystyle r \) (measured from Earth's center, not its surface) around Earth of mass \(\displaystyle M \), gravity itself supplies the centripetal force:\[\frac{GMm}{r^2} = \frac{mv^2}{r} \quad\Rightarrow\quad v^2 = \frac{GM}{r} \]So the orbital kinetic energy is\[KE = \frac{1}{2}mv^2 = \frac{GMm}{2r} \]Taking potential energy to be zero at infinite separation (the standard convention), the gravitational potential energy at radius \(\displaystyle r \) is\[PE = -\frac{GMm}{r} \]Adding these, the satellite's total mechanical energy in orbit is\[E = KE + PE = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r} \]This is negative — the satellite is bound to Earth, the same way an electron bound in an atom has negative energy. To free it completely (send it to infinity, arriving there with essentially zero leftover speed, the minimum condition for "escape"), its total energy must be raised to exactly zero. The energy you must supply is therefore the size of this negative total — its binding energy:\[\Delta E = 0 - E = \frac{GMm}{2r} \]Setting up the numbers. The orbital radius is measured from Earth's center, so you must add the height to the Earth's radius — using \(\displaystyle h \) alone here is the mistake to avoid:\[r = R + h = 6.4\times10^{6}\ \text{m} + 4.0\times10^{5}\ \text{m} = 6.8\times10^{6}\ \text{m} \]with \(\displaystyle m = 200\ \text{kg} \), \(\displaystyle M = 6.0\times10^{24}\ \text{kg} \), \(\displaystyle G = 6.67\times10^{-11}\ \text{N}\,\text{m}^2\,\text{kg}^{-2} \).Substituting:\[\Delta E = \frac{GMm}{2r} = \frac{(6.67\times10^{-11})(6.0\times10^{24})(200)}{2 \times 6.8\times10^{6}} \]Work the numerator first:\[GM = 6.67\times10^{-11} \times 6.0\times10^{24} = 4.002\times10^{14} \] \[GMm = 4.002\times10^{14} \times 200 = 8.004\times10^{16} \]and the denominator:\[2r = 2 \times 6.8\times10^{6} = 1.36\times10^{7}\ \text{m} \]so\[\Delta E = \frac{8.004\times10^{16}}{1.36\times10^{7}}\ \text{J} = 5.885\times10^{9}\ \text{J} \]The given data (\(\displaystyle M\), \(\displaystyle R\), \(\displaystyle h\)) carry only two significant figures, so the result should not carry more precision than that — round the final value only, not the intermediate steps:\[\Delta E \approx 5.9\times10^{9}\ \text{J} \]This is a huge amount of energy (about $\displaystyle 5.9$ gigajoules) — it has to be, since it must undo both the satellite's binding to Earth's gravity and account for the kinetic energy it already carries in orbit.Answer: The satellite must be given about \(\displaystyle 5.9\times10^{9}\ \text{J}\) of energy — its orbital binding energy \(\displaystyle \dfrac{GMm}{2r} \), with \(\displaystyle r = R+h = 6.8\times10^{6}\ \text{m}\) — to leave Earth's gravitational influence.

Two stars falling together convert gravitational potential energy into kinetic energy — and because they start from rest, symmetry forces them to arrive with equal speeds.Setting up: momentum tells you the two speeds are equalNo external force acts on this two-star system (only their mutual gravity, an internal force), so the total momentum is conserved. Since both stars start at rest, the total momentum is zero throughout the motion. If at some instant the stars (equal mass \(\displaystyle M\)) have speeds \(\displaystyle v_1\) and \(\displaystyle v_2\) directed toward each other, \[Mv_1 - Mv_2 = 0 \implies v_1 = v_2 = v. \] So right up to collision, each star moves at the same speed \(\displaystyle v\) — this is what lets you write the kinetic energy of the pair as \(\displaystyle Mv^2\) instead of tracking two different speeds.Setting up: energy conservationGravity is a conservative force, so the total mechanical energy (kinetic + gravitational potential) of the two-star system is conserved. For two masses \(\displaystyle M\) separated by a center-to-center distance \(\displaystyle r\), the gravitational potential energy is \[U = -\frac{GM^2}{r}, \] where \(\displaystyle G = 6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2}\) is the universal gravitational constant — not to be confused with \(\displaystyle g\), which is a local surface value and does not appear anywhere in this calculation.Initial state: separation \(\displaystyle r_i = 10^9\ \text{km} = 10^{12}\ \text{m}\) (center to center), speeds negligible, so \(\displaystyle KE_i \approx 0\).Final state (just at collision): the surfaces touch, so the centers are separated by the sum of the two radii, \[r_f = R + R = 2R = 2\times10^4\ \text{km} = 2\times10^7\ \text{m}, \] and each star moves at speed \(\displaystyle v\), giving \(\displaystyle KE_f = \frac{1}{2}Mv^2 + \frac{1}{2}Mv^2 = Mv^2\).Solving for vConservation of energy, \(\displaystyle KE_i + U_i = KE_f + U_f\): \[0 - \frac{GM^2}{r_i} = Mv^2 - \frac{GM^2}{r_f} \] \[Mv^2 = GM^2\left(\frac{1}{r_f} - \frac{1}{r_i}\right) \] \[v^2 = GM\left(\frac{1}{2R} - \frac{1}{r_i}\right) \]Substituting the numbersWith \(\displaystyle M = 2\times10^{30}\ \text{kg}\), \[GM = (6.67\times10^{-11})(2\times10^{30}) = 1.334\times10^{20}\ \text{m}^3\text{s}^{-2}. \]The two reciprocal terms: \[\frac{1}{2R} = \frac{1}{2\times10^7\ \text{m}} = 5.0\times10^{-8}\ \text{m}^{-1}, \qquad \frac{1}{r_i} = \frac{1}{10^{12}\ \text{m}} = 1.0\times10^{-12}\ \text{m}^{-1}. \] The second term is four orders of magnitude smaller than the first, so it does not shift the answer at the precision this problem calls for — the stars start so far apart that their initial potential energy is essentially zero compared with what they have gained by the time they touch: \[v^2 \approx (1.334\times10^{20}\ \text{m}^3\text{s}^{-2})(5.0\times10^{-8}\ \text{m}^{-1}) = 6.67\times10^{12}\ \text{m}^2\text{s}^{-2}. \] \[v = \sqrt{6.67\times10^{12}}\ \text{m/s} = 2.582\times10^6\ \text{m/s}. \]Rounding to three significant figures — matching \(\displaystyle G\), the least-precisely-known quantity used in the calculation — gives \[v \approx 2.58\times10^6\ \text{m/s}. \]DirectionEach star moves along the line joining the two centers, straight toward the other star — that is the only direction gravity can pull it in this problem, and it is also the direction fixed by the initial head-on approach.Answer: Each star is moving at about \(\displaystyle 2.58\times10^{6}\ \text{m/s}\) (roughly $\displaystyle 2580$ km/s), directed straight at the other star along the line joining their centers, at the instant of collision.

A shell of matter attracts everything outside it exactly as if all its mass were squeezed to a point at its centre — that is what lets you treat each $\displaystyle 100$ kg sphere as a point mass once you are outside its surface.Given: The midpoint of the line joining the centres is at a distance \[r = \frac{d}{2} = \frac{1.0 \, \text{m}}{2} = 0.50 \, \text{m} \] from each centre. Since \(\displaystyle r = 0.50 \, \text{m} \) is well outside the surface of either sphere (\(\displaystyle R = 0.10 \, \text{m} \)), Newton's shell theorem lets each sphere be replaced by a point mass \(\displaystyle M \) sitting at its own centre.Force at the midpointBy Newton's law of gravitation, a point mass \(\displaystyle M \) sets up a field of magnitude \[g = \frac{GM}{r^2} \] directed towards it, where \(\displaystyle G = 6.67 \times 10^{-11} \, \text{N m}^2\,\text{kg}^{-2} \) is the universal gravitational constant.At the midpoint, sphere A pulls a test mass toward A, and sphere B pulls it toward B, with equal magnitude because both centres are the same distance \(\displaystyle r = 0.50 \, \text{m} \) away: \[g_A = g_B = \frac{GM}{r^2} = \frac{(6.67\times10^{-11})(100)}{(0.50)^2} = 2.668 \times 10^{-8} \, \text{N kg}^{-1} \]These two pulls point in exactly opposite directions along the line joining the centres, so as vectors they cancel: \[g_{\text{net}} = g_A - g_B = 0 \]The gravitational field — and hence the force on any object placed there — is zero. Force is a vector: equal magnitudes pointing opposite ways cancel completely, even though each sphere individually is pulling hard.Potential at the midpointGravitational potential is a scalar, and scalars from different sources simply add — they do not cancel the way the forces did. The potential due to a point mass at distance \(\displaystyle r \) is \[V = -\frac{GM}{r} \] (always negative, since gravity is attractive). Adding the contributions of both spheres, \[V_{\text{net}} = -\frac{GM}{r} - \frac{GM}{r} = -\frac{2GM}{r} \]Substituting the values: \[V_{\text{net}} = -\frac{2(6.67\times10^{-11})(100)}{0.50} = -2.668 \times 10^{-8} \, \text{J kg}^{-1} \]The data (\(\displaystyle 1.0\,\text{m} \), \(\displaystyle 0.10\,\text{m} \)) is given to two significant figures, so the potential rounds to \[V_{\text{net}} \approx -2.7 \times 10^{-8} \, \text{J kg}^{-1} \]Is the object in equilibrium, and of what kind?Since the net force on an object placed at the midpoint is zero, it is in equilibrium there.To classify the equilibrium, displace the object by a small distance \(\displaystyle x \) along the line joining the centres, toward sphere A. Its distance from A becomes \(\displaystyle (0.50 - x) \) and from B becomes \(\displaystyle (0.50 + x) \). Because the pull of each sphere falls off as \(\displaystyle 1/r^2 \), the pull from the now-closer sphere A grows while the pull from the now-farther sphere B shrinks. The net force is no longer zero, and it points the same way as the displacement — toward A — pushing the object further from the midpoint instead of pulling it back. There is no restoring force along this line, so the equilibrium fails there.(A small sideways displacement, perpendicular to the line joining the centres, would be pulled back toward the line by the combined pull of the two spheres — that direction is locally stable — but a genuine equilibrium point must be stable in every direction. Since it is not stable along the line joining the centres, the midpoint as a whole is a point of unstable equilibrium.)Answer: The net gravitational force at the midpoint is zero (the two pulls are equal and opposite); the net gravitational potential there is \(\displaystyle \approx -2.7\times10^{-8} \, \text{J kg}^{-1} \). An object placed at the midpoint is in equilibrium, but the equilibrium is unstable — a small displacement along the line joining the centres grows instead of being restored.