NCERT Solutions · Class 11 Physics Gravitation

(a) Gravity cannot be shielded, because there is only one sign of "gravitational charge."Electrical shielding inside a hollow conductor works because a conductor is full of free charges of both signs. An external field pushes these charges around until the induced surface charge cancels the field everywhere inside — that cancellation is only possible because positive and negative charge both exist and can be separated.Mass is not like that. Newton's law of gravitation, \[F = \frac{Gm_1m_2}{r^2}, \] is always attractive, and there is no such thing as negative mass. A hollow sphere has no "opposite" mass it can push to the surface to cancel an external gravitational pull, so nothing analogous to electrical shielding is available. No arrangement of ordinary matter — a hollow sphere or anything else — can block gravity. A body inside a hollow sphere still feels the gravitational pull of everything outside it exactly as if the shell were not there.(b) A small ship cannot show gravity because it is in free fall; a very large station can, through the difference (tidal) in gravity across its size.An orbiting spaceship is in free fall: the whole ship and everything in it — the walls, the astronaut, a dropped pen — accelerate toward the Earth at the same local value of \(\displaystyle g = \dfrac{GM_e}{r^2}\), where \(\displaystyle M_e\) is Earth's mass and \(\displaystyle r\) the ship's distance from Earth's centre. Since there is no relative acceleration between the astronaut and the cabin, there is no contact force needed to hold him up, and that lack of contact force is what "weightlessness" means. If the ship is small, \(\displaystyle g\) has essentially the same magnitude and the same direction (toward Earth's centre) at every point inside it, so nothing inside reveals that gravity is even present.If the station is very large, this stops being true. Because \(\displaystyle g\) falls off as \(\displaystyle 1/r^2\), a point closer to Earth is pulled a little harder than a point farther away; and because \(\displaystyle g\) always points toward Earth's centre, the direction of the pull is slightly different at opposite ends of a large station (the lines converge). These small differences are exactly the differential — or tidal — effect of gravity, and they show up as a relative drift or stress between separated parts of the station (or between test masses placed at different points in it). So a large enough station lets the astronaut detect gravity, not through its ordinary pull (which free fall cancels), but through this tidal variation across the station's size.(c) The tide-raising effect depends on \(\displaystyle 1/d^{3}\), not \(\displaystyle 1/d^{2}\), and the Moon's closeness beats the Sun's much larger mass.Ordinary gravitational attraction on the whole Earth from a distant mass \(\displaystyle M\) at distance \(\displaystyle d\) is \[F = \frac{GM_e M}{d^2}. \] Tides, however, are not caused by this force itself but by how much it differs between the near side and the far side of the Earth (radius \(\displaystyle R_e\)), since it is that difference which stretches the Earth and its oceans. The pull on the near face is \(\displaystyle \dfrac{GM}{(d-R_e)^2}\) and on the far face is \(\displaystyle \dfrac{GM}{(d+R_e)^2}\). For \(\displaystyle R_e \ll d\), the difference is well approximated by the derivative of \(\displaystyle g(d)=GM/d^2\): \[\Delta g \approx \left|\frac{dg}{dd}\right|(2R_e) = \frac{2GM}{d^3}(2R_e) = \frac{4GM R_e}{d^3}. \] So the tide-raising effect scales as \(\displaystyle M/d^{3}\) — one extra power of \(\displaystyle d\) in the denominator compared with the force itself, which scales as \(\displaystyle M/d^{2}\).Using \(\displaystyle M_{\text{sun}} = 2\times10^{30}\ \text{kg}\), \(\displaystyle d_{\text{sun}} = 1.5\times10^{11}\ \text{m}\), \(\displaystyle M_{\text{moon}} = 7.36\times10^{22}\ \text{kg}\), \(\displaystyle d_{\text{moon}} = 3.84\times10^{8}\ \text{m}\) (the data given in the exercises that follow):Ordinary force ratio: \[\frac{F_{\text{sun}}}{F_{\text{moon}}} = \frac{M_{\text{sun}}}{M_{\text{moon}}}\left(\frac{d_{\text{moon}}}{d_{\text{sun}}}\right)^2 = (2.72\times10^{7})(2.56\times10^{-3})^2 = (2.72\times10^{7})(6.55\times10^{-6}) \approx 1.8\times10^2. \] The Sun's pull on the Earth is about $\displaystyle 180$ times the Moon's pull — confirming the statement in the question.Tidal ratio: \[\frac{T_{\text{sun}}}{T_{\text{moon}}} = \frac{M_{\text{sun}}}{M_{\text{moon}}}\left(\frac{d_{\text{moon}}}{d_{\text{sun}}}\right)^3 = (2.72\times10^{7})(2.56\times10^{-3})^3 = (2.72\times10^{7})(1.68\times10^{-8}) \approx 0.46. \] So \(\displaystyle \dfrac{T_{\text{moon}}}{T_{\text{sun}}} = \dfrac{1}{0.46} \approx 2.2\).The one extra power of distance in the tidal formula is decisive: the Moon is roughly $\displaystyle 390$ times closer to Earth than the Sun is, and \(\displaystyle (390)^3 \gg (390)^2\). That extra factor of closeness outweighs the Sun's much larger mass when it comes to the gradient of the field, even though it does not outweigh it for the field itself. This is the everyday reason ocean tides track the Moon's phase far more closely than the Sun's position.Answer: (a) No — gravity has no shielding because mass has only one sign, unlike charge, so nothing can be arranged to cancel an external gravitational pull. (b) A small ship cannot detect gravity, since free fall makes \(\displaystyle g\) locally uniform; a sufficiently large station can, through the tidal (differential) variation of \(\displaystyle g\) across its size. (c) Because the tide-raising effect varies as \(\displaystyle 1/d^3\) while the force itself varies as \(\displaystyle 1/d^2\), the Moon's much smaller distance makes its tidal effect about $\displaystyle 2.2$ times the Sun's, even though the Sun's direct pull on Earth is about $\displaystyle 180$ times the Moon's.

Gravity is not constant — it changes with both height above and depth below Earth's surface, and only mass of the body drops out, not mass of the Earth. Work through each part using the standard formulas for \(\displaystyle g\) at height, at depth, and the potential-energy difference.(a) Acceleration due to gravity decreases with increasing altitude.At a height \(\displaystyle h\) above the surface, treating the Earth as a uniform sphere of mass \(\displaystyle M\) and radius \(\displaystyle R\), \[g(h) = \frac{GM}{(R+h)^2}. \] As \(\displaystyle h\) increases, the denominator \(\displaystyle (R+h)^2\) grows, so \(\displaystyle g(h)\) falls. For \(\displaystyle h \ll R\) this is often written to first order as \[g(h) \approx g\left(1 - \frac{2h}{R}\right), \] which makes the decrease with \(\displaystyle h\) explicit. Either way, going up weakens gravity — this is why astronauts in low orbit still feel most of Earth's pull, but slightly less than someone standing on the ground.(b) Acceleration due to gravity decreases with increasing depth.Below the surface, only the mass enclosed within radius \(\displaystyle r\) (measured from the centre) contributes to gravity, and for a sphere of uniform density that enclosed mass is \(\displaystyle M(r) = M\left(\dfrac{r}{R}\right)^{3}\). At depth \(\displaystyle d\), so \(\displaystyle r = R - d\), \[g(d) = \frac{GM(r)}{r^2} = \frac{GM}{R^3}\,r = g\left(1 - \frac{d}{R}\right). \] As \(\displaystyle d\) increases, \(\displaystyle g(d)\) decreases linearly, reaching zero at the centre (\(\displaystyle d=R\)) — a common point of confusion is expecting gravity to be strongest at the centre, but the mass pulling you down keeps shrinking as you dig in, and it wins out.(c) Acceleration due to gravity is independent of the mass of the body, but depends on the mass of the Earth.From \(\displaystyle g = \dfrac{GM}{R^2}\), the mass \(\displaystyle m\) of the falling body never appears — this is exactly the statement of the equivalence of inertial and gravitational mass, and it is why a feather and a hammer fall at the same rate in vacuum. But \(\displaystyle M\), the Earth's mass, sits directly in the formula: a planet with a different mass would give every object on it a different \(\displaystyle g\).(d) The formula \(\displaystyle -GMm\left(\dfrac{1}{r_2}-\dfrac{1}{r_1}\right)\) is more accurate than \(\displaystyle mg(r_2-r_1)\).The gravitational potential energy of a mass \(\displaystyle m\) at distance \(\displaystyle r\) from Earth's centre (mass \(\displaystyle M\)) is exactly \[U(r) = -\frac{GMm}{r}. \] So the exact difference in potential energy between two points at distances \(\displaystyle r_2\) and \(\displaystyle r_1\) from the centre is \[U(r_2) - U(r_1) = -\frac{GMm}{r_2} - \left(-\frac{GMm}{r_1}\right) = -GMm\left(\frac{1}{r_2}-\frac{1}{r_1}\right), \] and this holds at any separation, near the surface or far from it.The formula \(\displaystyle mg(r_2-r_1)\) instead assumes \(\displaystyle g\) is the same constant value at both points — valid only when \(\displaystyle r_2 - r_1 \ll R\), i.e., both points lie close to the surface where \(\displaystyle g\) barely changes. The moment the two points are far apart (or one is at a large altitude), this approximation breaks down because \(\displaystyle g\) itself varies with \(\displaystyle r\) as shown in part (a). The exact inverse-\(\displaystyle r\) formula makes no such assumption, so it remains valid everywhere and is therefore the more accurate expression.Answer: (a) decreases; (b) decreases; (c) mass of the body (independent of \(\displaystyle m\)), but depends on mass of the earth \(\displaystyle M\); (d) \(\displaystyle -GMm\left(\dfrac{1}{r_2}-\dfrac{1}{r_1}\right)\) is more accurate.

A faster orbit is not a bigger one — Kepler's third law ties the orbital period and the orbital size together, and speeding up the trip around the Sun means shrinking the orbit, not growing it.Kepler's third law of planetary motion states that for any planet going around the Sun, the square of its orbital period \(\displaystyle T \) is proportional to the cube of its orbital radius \(\displaystyle r \) (its orbital size):\[T^2 \propto r^3 \]Here \(\displaystyle T \) is the time the planet takes to complete one revolution around the Sun, and \(\displaystyle r \) is the radius of its (assumed circular) orbit.For two planets, this proportionality gives\[\left(\frac{T'}{T}\right)^2 = \left(\frac{r'}{r}\right)^3 \]where \(\displaystyle T, r \) belong to Earth and \(\displaystyle T', r' \) belong to the new planet.Reading "twice as fast" correctly. Going around twice as fast means completing one revolution in half the time — the new planet's period is half Earth's period, not its speed of revolution in the everyday sense. So\[T' = \frac{T}{2} \quad\Rightarrow\quad \frac{T'}{T} = \frac{1}{2} \]Substituting into the law:\[\left(\frac{1}{2}\right)^2 = \left(\frac{r'}{r}\right)^3 \]\[\frac{1}{4} = \left(\frac{r'}{r}\right)^3 \]Solving for the radius ratio:\[\frac{r'}{r} = \left(\frac{1}{4}\right)^{1/3} = 4^{-1/3} \]Evaluating numerically, \(\displaystyle 4^{1/3} = 1.587\ldots \), so\[\frac{r'}{r} = \frac{1}{1.587} = 0.6300\ldots \]"Twice" is an exact factor here, so the ratio is properly stated to two significant figures: \(\displaystyle r'/r \approx 0.63 \).This confirms the physical picture: a planet that completes its orbit twice as fast must be running on a smaller track — its orbital radius is only about $\displaystyle 63$% of Earth's, not larger. A common slip is to assume "faster" means "farther out"; it is the opposite, because a planet closer to the Sun feels a stronger gravitational pull and sweeps around in less time.Answer: The planet's orbital radius would be about \(\displaystyle 0.63 \) times that of the Earth (i.e., \(\displaystyle r' = (1/4)^{1/3}\,r_E \approx 0.63\,r_E \)) — a smaller orbit, not a larger one.

Kepler's third law lets you weigh Jupiter using one of its moons, without ever needing to know \(\displaystyle G\).For a body of mass \(\displaystyle m\) moving in a circular orbit of radius \(\displaystyle R\) and period \(\displaystyle T\) around a much larger central mass \(\displaystyle M\), gravity supplies the centripetal force: \[\frac{GMm}{R^2} = \frac{4\pi^2 m R}{T^2} \]The orbiting body's own mass \(\displaystyle m\) cancels from both sides — this is exactly why the same relation works whether the "satellite" is a moon circling a planet or a planet circling the Sun. Solving for the central mass gives Kepler's third law in the form you need: \[M = \frac{4\pi^2 R^3}{G T^2} \]Apply this once to Io orbiting Jupiter, and once to Earth orbiting the Sun: \[M_J = \frac{4\pi^2 R_{Io}^3}{G\,T_{Io}^2}, \qquad M_S = \frac{4\pi^2 R_E^3}{G\,T_E^2} \]Here \(\displaystyle R_{Io}=4.22\times10^{8}\ \text{m}\) and \(\displaystyle T_{Io}=1.769\) days are the Io data given in the problem; \(\displaystyle R_E = 1.496\times10^{11}\ \text{m}\) ($\displaystyle 1$ astronomical unit) and \(\displaystyle T_E = 365.25\) days are the standard Earth–Sun values.Divide the two equations. \(\displaystyle 4\pi^2\) and \(\displaystyle G\) are the same constants in both — they were never specific to Jupiter or the Sun — so they cancel exactly: \[\frac{M_J}{M_S} = \left(\frac{R_{Io}}{R_E}\right)^{3}\left(\frac{T_E}{T_{Io}}\right)^{2} \]This cancellation is the whole trick of the method: because only a ratio of periods appears, both \(\displaystyle T_{Io}\) and \(\displaystyle T_E\) can stay in days — there is no need to convert either to seconds, and no need to look up \(\displaystyle G\) at all.Radius ratio \[\frac{R_{Io}}{R_E} = \frac{4.22\times10^{8}\ \text{m}}{1.496\times10^{11}\ \text{m}} = 2.821\times10^{-3} \] \[\left(\frac{R_{Io}}{R_E}\right)^{3} = (2.821\times10^{-3})^{3} = 2.245\times10^{-8} \]Period ratio \[\frac{T_E}{T_{Io}} = \frac{365.25\ \text{d}}{1.769\ \text{d}} = 206.5 \] \[\left(\frac{T_E}{T_{Io}}\right)^{2} = (206.5)^{2} = 4.264\times10^{4} \]Combine \[\frac{M_J}{M_S} = (2.245\times10^{-8})(4.264\times10^{4}) = 9.57\times10^{-4} \]Taking the reciprocal, \(\displaystyle 1/(9.57\times10^{-4}) \approx 1045\) — so Jupiter's mass comes out to about \(\displaystyle 1/1000\) of the Sun's mass, exactly the ratio the question asks you to demonstrate.As a check by the direct route: converting \(\displaystyle T_{Io}=1.769\) days \(\displaystyle =1.528\times10^{5}\ \text{s}\) and substituting into \(\displaystyle M_J = 4\pi^2R_{Io}^3/(GT_{Io}^2)\) with \(\displaystyle G=6.674\times10^{-11}\ \text{N m}^2\text{kg}^{-2}\) gives \(\displaystyle M_J \approx 1.90\times10^{27}\ \text{kg}\). Dividing by the Sun's known mass \(\displaystyle M_S = 1.99\times10^{30}\ \text{kg}\) gives \(\displaystyle 9.6\times10^{-4}\) again — the same result by a second, independent path.\(\displaystyle R_{Io}\) is given to three significant figures, which limits the final ratio to three figures as well.Answer: \(\displaystyle M_J/M_S \approx 9.57\times10^{-4}\) — Jupiter's mass is about one-thousandth (\(\displaystyle \approx 1/1045\)) the mass of the Sun.

Model the whole galaxy as a single central mass, and treat the star exactly like a planet orbiting the Sun.The diameter of the Milky Way is given as \(\displaystyle 10^{5}\) ly, so its radius is \(\displaystyle 5\times10^{4}\) ly — exactly the distance of this star from the galactic centre. That means essentially the entire mass of the galaxy lies inside the star's orbit, so by the shell theorem (the same rule that says the gravity of a uniform sphere acts, from outside, as if all its mass sat at the centre) that mass pulls on the star as a single point mass \(\displaystyle M\) at the centre. This lets you reuse Kepler's third law with \(\displaystyle M\) playing the role of the "Sun."Newton's law of gravitation supplies the centripetal force.For a star of mass \(\displaystyle m\) moving in a circle of radius \(\displaystyle r\) with period \(\displaystyle T\), angular speed \(\displaystyle \omega = 2\pi/T\), gravity provides the centripetal force: \[\frac{GMm}{r^{2}} = m\omega^{2}r = m\left(\frac{2\pi}{T}\right)^{2}r \] Cancel \(\displaystyle m\) (the star's own mass doesn't matter) and solve for \(\displaystyle T\): \[T = 2\pi\sqrt{\frac{r^{3}}{GM}} \] Here \(\displaystyle G = 6.67\times10^{-11}\ \text{N m}^{2}\text{kg}^{-2}\) is the universal gravitational constant — not to be confused with \(\displaystyle g\), the acceleration due to gravity at a planet's surface, which doesn't appear anywhere in this problem.Find the enclosed mass \(\displaystyle M\).Number of stars \(\displaystyle N = 2.5\times10^{11}\), each of one solar mass. Using the standard value \(\displaystyle M_{\odot} = 2\times10^{30}\ \text{kg}\): \[M = N M_{\odot} = (2.5\times10^{11})(2\times10^{30}\ \text{kg}) = 5\times10^{41}\ \text{kg} \]Convert the orbital radius to metres.\(\displaystyle r = 50{,}000\ \text{ly} = 5\times10^{4}\ \text{ly}\). Using \(\displaystyle 1\ \text{ly} = 9.46\times10^{15}\ \text{m}\): \[r = (5\times10^{4})(9.46\times10^{15}\ \text{m}) = 4.73\times10^{20}\ \text{m} \] (A common slip here is plugging in the diameter, \(\displaystyle 10^{5}\) ly, instead of the radius — the diameter is only given so you can check that the star sits right at the edge of the galaxy, which is what justifies treating all the mass as "enclosed.")Substitute into the period formula. \[r^{3} = (4.73\times10^{20}\ \text{m})^{3} = 1.058\times10^{62}\ \text{m}^{3} \] \[GM = (6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2})(5\times10^{41}\ \text{kg}) = 3.34\times10^{31}\ \text{N m}^2\text{kg}^{-1} \] \[\frac{r^{3}}{GM} = \frac{1.058\times10^{62}}{3.34\times10^{31}} = 3.17\times10^{30}\ \text{s}^{2} \] \[T = 2\pi\sqrt{3.17\times10^{30}\ \text{s}^{2}} = 2\pi\,(1.78\times10^{15}\ \text{s}) = 1.12\times10^{16}\ \text{s} \]Convert to years — a period of \(\displaystyle 10^{16}\) seconds means nothing on a human or even astronomical scale until it's converted. One year \(\displaystyle \approx 3.16\times10^{7}\ \text{s}\): \[T = \frac{1.12\times10^{16}\ \text{s}}{3.16\times10^{7}\ \text{s/yr}} \approx 3.5\times10^{8}\ \text{yr} \]The data given (\(\displaystyle 2.5\times10^{11}\) stars, \(\displaystyle 50{,}000\) ly, \(\displaystyle 10^{5}\) ly) all carry two significant figures, so the result is rounded to two significant figures rather than carried through at full calculator precision.Answer: \(\displaystyle T \approx 3.5\times10^{8}\) years (about $\displaystyle 355$ million years) — the star takes roughly $\displaystyle 350$ million years to complete one revolution about the galactic centre.

A satellite in orbit is already moving, so it already owns some kinetic energy — that head start is exactly why it takes less extra energy to escape than a resting object at the same height, and exactly why its total energy comes out negative of its kinetic energy rather than its potential energy.(a) Total energy is negative of the kinetic energyTake a satellite of mass \(\displaystyle m \) moving in a circular orbit of radius \(\displaystyle r \) around the Earth (mass \(\displaystyle M \)). Gravity supplies the centripetal force that holds it in that orbit, so by Newton's law of gravitation,\[\frac{GMm}{r^2} = \frac{mv^2}{r} \]where \(\displaystyle G \) is the universal gravitational constant and \(\displaystyle v \) is the orbital speed. Cancelling \(\displaystyle m \) and one factor of \(\displaystyle r \),\[v^2 = \frac{GM}{r} \]The kinetic energy of the satellite is therefore\[KE = \frac{1}{2}mv^2 = \frac{GMm}{2r} \]With the zero of potential energy fixed at infinity (the convention this question specifies), the gravitational potential energy at radius \(\displaystyle r \) is\[PE = -\frac{GMm}{r} \]The minus sign is not optional bookkeeping — it says the satellite is bound, sitting in a potential well relative to "free at infinity."Adding the two gives the total mechanical energy of the orbit:\[E = KE + PE = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r} \]Now compare this with the kinetic energy computed above: \(\displaystyle KE = \dfrac{GMm}{2r} \), so\[E = -\frac{GMm}{2r} = -KE \]The total energy is the negative of the kinetic energy. (It also happens to equal \(\displaystyle PE/2 \), not \(\displaystyle -PE \) — that second relation is the one people confuse it with, so it's worth keeping the two straight: \(\displaystyle E = -KE = \tfrac{1}{2}PE \).)(b) Less energy is needed for the orbiting satellite"Escaping Earth's influence" means being carried out to \(\displaystyle r \to \infty \) with just enough energy left to arrive there at rest, i.e. reaching total mechanical energy \(\displaystyle E = 0 \). The energy that must be added to a body is therefore the amount needed to raise its current total energy up to zero, which is just \(\displaystyle -E \) (the size of the negative total energy it starts with).For the orbiting satellite, from part (a):\[E_{\text{satellite}} = -\frac{GMm}{2r} \quad\Rightarrow\quad \Delta E_{\text{satellite}} = \frac{GMm}{2r} \]For a stationary object of the same mass \(\displaystyle m \), sitting at rest at the same height \(\displaystyle r \), the kinetic energy is zero, so its total energy is just its potential energy:\[E_{\text{stationary}} = 0 + \left(-\frac{GMm}{r}\right) = -\frac{GMm}{r} \quad\Rightarrow\quad \Delta E_{\text{stationary}} = \frac{GMm}{r} \]Comparing the two required amounts,\[\Delta E_{\text{satellite}} = \frac{GMm}{2r} < \frac{GMm}{r} = \Delta E_{\text{stationary}} \]The satellite needs only half as much extra energy as the stationary object. This makes physical sense: the satellite is already moving, so part of the energy needed to escape (the kinetic-energy part) has already been supplied by putting it into orbit in the first place — only the potential-energy deficit is left to overcome, and even that is only half the size because the orbital kinetic energy offsets half of it.Answer: (a) The total energy of the orbiting satellite is negative of its kinetic energy (\(\displaystyle E = -KE\)). (b) The energy needed to launch the satellite out of Earth's gravitational influence is less than the energy needed to send a stationary object at the same height out of Earth's influence.

Escape speed comes from energy conservation, and only the quantities that actually survive that conservation law can affect it.Escape speed is the minimum speed a body needs, moving away from the earth, so that it never falls back — it just barely reaches infinity with zero kinetic energy left.Use the law of conservation of mechanical energy. Take a body of mass \(\displaystyle m\) launched from a point at distance \(\displaystyle R\) from the earth's centre (its speed there is \(\displaystyle v_e\)): At infinity, for the minimum escape speed, both the kinetic energy and the potential energy become zero. Total energy is conserved, so the total energy at launch must equal zero too:\[\frac{1}{2}mv_e^2 - \frac{GMm}{R} = 0 \]The mass \(\displaystyle m\) of the body appears in both terms and cancels:\[v_e = \sqrt{\frac{2GM}{R}} \]This is the formula to check each dependence against.(a) Mass of the body — no. The body's mass \(\displaystyle m\) cancelled out of the equation above. A pebble and a satellite launched from the same point need exactly the same escape speed (air resistance aside, which this idealised treatment ignores).(b) Location from where it is projected — no. \(\displaystyle v_e\) depends only on \(\displaystyle R\), the distance from the earth's centre. Treating the earth as a uniform sphere, every point on its surface is the same distance \(\displaystyle R_E\) from the centre, so the escape speed is the same whether the body is launched from a pole, the equator, or anywhere else at sea level. (In reality the earth bulges slightly at the equator, but that small variation in \(\displaystyle R\) is outside what this idealised model accounts for.)(c) Direction of projection — no. Energy conservation is a statement about the scalar quantities \(\displaystyle \frac{1}{2}mv^2\) and \(\displaystyle -\dfrac{GMm}{r}\) — neither depends on which way the velocity vector points. So the speed needed to escape is the same whether the body is fired straight up or at an angle, as long as its path doesn't run into the earth itself or through resistance that depends on the path taken.(d) Height of the location from where the body is launched — yes. This is the one case where the formula's variable actually changes. Launching from a height \(\displaystyle h\) above the surface changes the starting distance from the centre to \(\displaystyle R_E + h\), so\[v_e = \sqrt{\frac{2GM}{R_E+h}} \]Since \(\displaystyle R_E+h\) grows with \(\displaystyle h\), \(\displaystyle v_e\) gets smaller the higher up the body starts — which is exactly why launching from a tall tower, or better yet from orbit, takes less speed to escape than launching from the ground.Answer: Escape speed does not depend on the mass of the body (a), the location on the earth's surface it is projected from (b), or the direction of projection (c). It does depend on the height above the surface from which it is launched (d), since \(\displaystyle v_e=\sqrt{2GM/(R_E+h)}\) decreases as \(\displaystyle h\) increases.

The Sun's gravity pulls on the comet with a force that is always directed straight at the Sun (a central force), and this single fact answers all six parts at once. A central force cannot produce any torque about the Sun, so angular momentum is conserved throughout the orbit. Gravity is also a conservative force, so total mechanical energy is conserved. But the comet's distance from the Sun changes enormously along a highly elliptical orbit — swinging from very close at perihelion to very far at aphelion — and that changing distance is what makes speed, angular speed, kinetic energy, and potential energy all vary.(a) Linear speed — not constant. By Kepler's second law, the line joining the comet to the Sun sweeps out equal areas in equal times. Near perihelion (closest approach) the comet is close to the Sun, so to sweep the same area in the same time it must move through a much longer arc — it moves fast. Near aphelion (farthest point) it is far from the Sun, sweeps the same area over a short arc, and moves slowly. So the linear (orbital) speed is largest at perihelion and smallest at aphelion — it is not constant.(b) Angular speed — not constant. Angular speed \(\displaystyle \omega = v/r \), where \(\displaystyle v\) is the linear speed and \(\displaystyle r\) is the comet's distance from the Sun. At perihelion both effects work the same way: \(\displaystyle v\) is large and \(\displaystyle r\) is small, so \(\displaystyle \omega\) is large. At aphelion \(\displaystyle v\) is small and \(\displaystyle r\) is large, so \(\displaystyle \omega\) is small. Angular speed therefore also varies along the orbit — it is not constant. (Do not confuse this with angular momentum below: angular speed changes even though angular momentum does not, because angular momentum depends on \(\displaystyle r^2\omega\), not on \(\displaystyle \omega\) alone.)(c) Angular momentum — constant. Angular momentum about the Sun is \(\displaystyle L = mvr\sin\theta\), taken about the Sun. The gravitational force on the comet always points along the line joining the comet to the Sun, so the torque of this force about the Sun, \[\vec{\tau} = \vec{r}\times\vec{F}, \] is zero at every instant (\(\displaystyle \vec r\) and \(\displaystyle \vec F\) are anti-parallel, so the cross product vanishes). Since \(\displaystyle \vec\tau = d\vec L/dt = 0\), the angular momentum \(\displaystyle \vec L\) stays constant throughout the orbit. This is exactly the physical content of Kepler's second law used in part (a) — equal areas in equal times is just conservation of angular momentum written geometrically.(d) Kinetic energy — not constant. Kinetic energy is \(\displaystyle KE = \tfrac12 mv^2\). Since the linear speed \(\displaystyle v\) is not constant (part a), \(\displaystyle KE\) is not constant either: it is maximum at perihelion (where \(\displaystyle v\) is largest) and minimum at aphelion (where \(\displaystyle v\) is smallest).(e) Potential energy — not constant. Gravitational potential energy of the comet–Sun system is \[U(r) = -\frac{GMm}{r}, \] where \(\displaystyle G\) is the universal gravitational constant, \(\displaystyle M\) is the Sun's mass, \(\displaystyle m\) is the comet's mass, and \(\displaystyle r\) is the comet–Sun separation. Because \(\displaystyle r\) changes continuously along an elliptical orbit, \(\displaystyle U\) changes too: it is most negative (lowest) at perihelion, where \(\displaystyle r\) is smallest, and least negative (highest) at aphelion, where \(\displaystyle r\) is largest. So potential energy is not constant.(f) Total energy — constant. The only force doing work on the comet is gravity, and gravity is a conservative force — a comet does not lose energy to friction or drag in empty space (and the problem tells us to neglect any mass loss, so no energy is carried away by ejected material either). For an isolated system acted on only by a conservative internal force, mechanical energy is conserved: \[E = KE + U = \text{constant}. \] So even though \(\displaystyle KE\) and \(\displaystyle U\) individually swing between large and small values as the comet moves from perihelion to aphelion, their sum stays exactly the same at every point of the orbit — the increase in kinetic energy as the comet falls toward the Sun is paid for exactly by the decrease (more negative) in potential energy, and vice versa as it recedes.Answer: (a) linear speed — not constant (maximum at perihelion, minimum at aphelion); (b) angular speed — not constant (same reason); (c) angular momentum — constant (gravity is a central force, so zero torque about the Sun); (d) kinetic energy — not constant (tracks the speed); (e) potential energy — not constant (tracks the distance from the Sun, \(\displaystyle U=-GMm/r\)); (f) total energy — constant (gravity is conservative, so \(\displaystyle KE+U\) is conserved throughout the orbit).

In space, gravity's pull on body fluids disappears, so fluid moves the opposite way from what happens on Earth — up towards the head, not down towards the feet.On Earth, gravity constantly pulls blood and other body fluids downward. The body's circulatory system works against this pull, and a certain amount of fluid naturally settles towards the lower limbs — this is why feet and ankles swell when gravity's downward pull on fluid is not adequately opposed (for example, after long periods of standing).In space, an astronaut is in a state of weightlessness — this is because the astronaut and the spacecraft are both in free fall around the Earth, so the astronaut experiences no effective gravitational force relative to the spacecraft. With that downward pull gone, the mechanism that normally keeps fluid pooled near the feet is absent. Instead:The step people get wrong here is treating "zero gravity" as simply "nothing changes." In fact weightlessness actively reverses the fluid-distribution pattern the body is used to on Earth, which is why the face swells while the feet do not — the opposite of what many expect at first.Answer: (b) swollen face, (c) headache, and (d) orientational problem are likely to afflict an astronaut in space; (a) swollen feet is not, since the absence of gravity removes the downward pull on body fluids that causes feet to swell on Earth.

A field that is forced by symmetry to lie along one line cannot then choose to be zero or sideways — trace the symmetry first, then use "gravity only pulls" to fix which way along that line it points.Call \(\displaystyle O\) the centre of curvature of the hemispherical shell — the same point that would be the centre of the full sphere if the other half of the shell were added back. \(\displaystyle O\) sits exactly in the plane of the shell's circular rim (its "flat face"), and all the mass of the shell lies to one side of that plane, curving away from \(\displaystyle O\).Step $\displaystyle 1$ — Symmetry fixes the line, and rules out arrow \(\displaystyle b\).Spin the shell about the axis that passes through \(\displaystyle O\) perpendicular to the flat face. Because the shell is uniform and this rotation carries the shell exactly onto itself, whatever gravitational intensity vector \(\displaystyle \vec{g}\) exists at \(\displaystyle O\) must also be unchanged by the rotation — it is built entirely out of the mass distribution, and the mass distribution hasn't moved.A vector that survives being rotated through every angle about an axis cannot have any component perpendicular to that axis: a perpendicular component would itself get swept around to a new direction each time, so the only way it can stay fixed is if it is zero. Only a component along the axis can be rotation-invariant.Arrow \(\displaystyle b\) lies in the plane of the flat face, i.e. perpendicular to this axis. Step $\displaystyle 1$ alone rules it out — no field along \(\displaystyle b\) can survive the rotation test.Step $\displaystyle 2$ — Why the answer isn't "$\displaystyle 0$" either.For a complete spherical shell, the symmetry is stronger: rotating about any axis through the centre carries the shell onto itself, and by the same argument a field surviving rotation about every possible axis has nowhere left to point — it must vanish. That is the shell theorem, and it is the fact that option (iv) is quietly borrowing.But a hemispherical shell only has the weaker, one-axis symmetry from Step 1. That symmetry kills the sideways component (arrow \(\displaystyle b\)) but says nothing about the component along the axis itself, because every bit of mass lies strictly on one side of \(\displaystyle O\) — there is no matching mass on the other side to cancel it. So along the axis, nothing forces the field to zero.(A quick check that this is consistent: split a full spherical shell into its upper and lower hemispherical halves through \(\displaystyle O\). Each half's field at \(\displaystyle O\) points along the same axis by Step $\displaystyle 1$, and by mirror symmetry the two are equal in magnitude and opposite in direction, so they cancel to give the whole shell's zero field — without either individual hemisphere's field being zero.)Step $\displaystyle 3$ — Newton's law of gravitation fixes which way along the axis.For any mass element \(\displaystyle dm\) of the shell, the intensity it produces at \(\displaystyle O\) is \[d\vec{g} = -\frac{G\,dm}{r^2}\,\hat r, \] where \(\displaystyle \hat r\) points from \(\displaystyle O\) toward \(\displaystyle dm\) and \(\displaystyle G\) is the universal gravitational constant. The minus sign is the whole content of "gravity is always attractive": \(\displaystyle d\vec g\) points from \(\displaystyle O\) toward the mass, never away from it.Every element of the hemispherical shell lies on the concave side of \(\displaystyle O\) — the side arrow \(\displaystyle c\) points into, toward the shell material — not on the side arrow \(\displaystyle a\) points toward (which is empty space, outside the bowl). Adding up all these attractive contributions \(\displaystyle d\vec g\), each one pointing into the shell, the resultant \(\displaystyle \vec g\) at \(\displaystyle O\) must also point into the shell.So the vector consistent with both the symmetry argument (must lie on the axis) and the attraction argument (must point toward the mass) is arrow \(\displaystyle c\).Answer: (iii) c — the gravitational intensity at the centre of the hemispherical shell points along the axis of symmetry, into the shell (toward its mass), i.e. in the direction of arrow c.