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NCERT Exemplar · Class 9 Science Gravitation

26 questions · 26 still being checked

Short Answer Questions 16–22 (part 3 of 4)

  1. Exercise 16

    What is the source of centripetal force that a planet requires to revolve around the Sun? On what factors does that force depend?

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    NCERT’s answer
    Gravitational force. This force depends on the product of the masses of the planet and sun and the distance between them.
    Sun's gravitational pull on the planet supplies the centripetal force, directed toward the Sun.\[F=\frac{GM_sM_p}{r^2} \]\(\displaystyle M_s,M_p\) = masses of Sun, planet; \(\displaystyle r\) = distance between them; \(\displaystyle G\) = gravitational constant.Answer: The force depends on the mass of the Sun, the mass of the planet, and the distance between them.

  2. Exercise 17

    On the earth, a stone is thrown from a height in a direction parallel to the earth’s surface while another stone is simultaneously dropped from the same height. Which stone would reach the ground first and why?

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    Both stones will take the same time to reach the ground because the two stones fall from the same height.
    Vertical and horizontal motions are independent; the thrown stone's sideways motion does not change its downward fall.For both stones the initial vertical velocity \(\displaystyle u_y=0\): \[h=\frac12 g t^2 \ \Rightarrow\ t=\sqrt{\frac{2h}{g}} \]Same \(\displaystyle h\) and same \(\displaystyle g\) give the same \(\displaystyle t\).Answer: Both stones reach the ground at the same time.

  3. Exercise 18

    Suppose gravity of earth suddenly becomes zero, then in which direction will the moon begin to move if no other celestial body affects it?

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    The moon will begin to move in a straight line in the direction in which it was moving at that instant because the circular motion of moon is due to centripetal force provided by the gravitational force of earth.
    Earth's gravity was supplying the centripetal force that continually turns the Moon's velocity into a circular path.NCERT_Solution_Class9_Science_Exemplar_Ch10_Q18\[F=0 \ \Rightarrow\ a=0 \ \Rightarrow\ \vec{v}=\text{constant} \]By Newton's first law the Moon keeps the velocity it has at that instant — along the tangent to its orbit.Answer: The Moon moves off in a straight line, along the tangent to its orbit at that instant.

  4. Exercise 19

    Identical packets are dropped from two aeroplanes, one above the equator and the other above the north pole, both at height h. Assuming all conditions are identical, will those packets take same time to reach the surface of earth. Justify your answer.

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    The value of ‘g’ at the equator of the earth is less than that at poles. Therefore, the packet falls slowly at equator in comparison to the poles. Thus, the packet will remain in air for longer time interval, when it is dropped at the equator.
    Earth is not a perfect sphere — it bulges at the equator, so the equatorial radius exceeds the polar radius.\[g=\frac{GM}{R^2} \ \Rightarrow\ R_{equator}>R_{pole} \ \Rightarrow\ g_{equator}<g_{pole} \]\[h=\frac12 g t^2 \ \Rightarrow\ t=\sqrt{\frac{2h}{g}} \]A smaller \(\displaystyle g\) gives a larger \(\displaystyle t\).Answer: No — the packet dropped above the pole reaches the surface first, since \(\displaystyle g\) is larger there.

  5. Exercise 20

    The weight of any person on the moon is about 1\displaystyle 1/6\displaystyle 6 times that on the earth. He can lift a mass of 15\displaystyle 15 kg on the earth. What will be the maximum mass, which can be lifted by the same force applied by the person on the moon?

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    \(\displaystyle g_{e}\) = g and \(\displaystyle g_{m}\) = g Force applied to lift a mass of $\displaystyle 15$ kg at the earth, F = m \(\displaystyle g_{e}\) = $\displaystyle 15$ \(\displaystyle g_{e}\) N Therefore, the mass lifted by the same force on the moon, m = m F g = $\displaystyle 15$ g g = $\displaystyle 90$ kg
    The same muscular force lifts different masses against different accelerations due to gravity.\[F=m_eg_e \]\[F=m_mg_m=m_m\left(\frac{g_e}{6}\right) \]\[m_eg_e=m_m\frac{g_e}{6} \ \Rightarrow\ m_m=6m_e=6\times15\,\text{kg}=90\,\text{kg} \]Answer: \(\displaystyle 90\,\text{kg}\).

  6. Exercise 21

    Calculate the average density of the earth in terms of g, G and R.

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    NCERT’s answer
    g = G M R or M = × g R G ⇒ Density D = mass volume = × × e g R G V (Where \(\displaystyle V_{e}\) is the volume of the earth ) or D = π × × g R G R = g GR π
    Find earth's mass from its surface gravity, then divide by its volume.\[g=\frac{GM}{R^2} \ \Rightarrow\ M=\frac{gR^2}{G} \]\[\rho=\frac{M}{V}=\frac{gR^2/G}{\frac43\pi R^3}=\frac{3g}{4\pi GR} \]Answer: \(\displaystyle \rho=\dfrac{3g}{4\pi GR}\).

  7. Exercise 22

    The earth is acted upon by gravitation of Sun, even though it does not fall into the Sun. Why?

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    NCERT’s answer
    The gravitational force is responsible for providing the necessary centripetal force. Long Answer Questions
    Earth has a tangential (orbital) velocity; the Sun's gravity continuously acts as the centripetal force, curving this velocity into a closed orbit instead of letting Earth fall straight in.NCERT_Solution_Class9_Science_Exemplar_Ch10_Q22\[\frac{GM_sM_e}{r^{2}}=\frac{M_ev^{2}}{r} \ \Rightarrow\ v=\sqrt{\frac{GM_s}{r}} \]Answer: The Sun's gravity supplies the centripetal force that keeps turning Earth's tangential velocity into an orbit, rather than pulling it straight into the Sun.