Gravitation

Practise Gravitation for FBISE Grade 9–10 Physics (Grade 9–10): 8 multiple-choice questions with instant checking and worked explanations. Free, no sign-up.

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1. Question 1

   A communication satellite is moved from a low Earth orbit to a geostationary orbit with a significantly larger orbital radius. How do the gravitational force exerted on the satellite by Earth and its orbital speed change?
   • ABoth gravitational force and orbital speed decrease.
   • BBoth gravitational force and orbital speed increase.
   • CGravitational force increases while orbital speed decreases.
   • DGravitational force decreases while orbital speed increases.
   Show answer and explanation

   Answer: A) Both gravitational force and orbital speed decrease.

   Both the gravitational attraction ($F\propto 1/r^2$) and the required orbital speed ($v\propto 1/\sqrt{r}$) fall as the orbital radius grows, so they each decrease when a satellite moves to a higher orbit. The other options incorrectly assume one or both quantities increase with distance.

2. Question 2

   An artificial satellite orbits Earth at an altitude equal to Earth's radius $R$, where the local gravitational acceleration is a fraction of its surface value $g$. If the satellite relies entirely on Earth's gravity to maintain its circular orbit, what is its orbital speed?
   • A$\sqrt{2gR}$
   • B$\sqrt{\frac{gR}{2}}$
   • C$\sqrt{gR}$
   • D$\frac{\sqrt{gR}}{2}$
   Show answer and explanation

   Answer: B) $\sqrt{\frac{gR}{2}}$

   At an altitude equal to Earth's radius the orbital radius is $2R$. Using $v=\sqrt{GM/r}$ and $g=GM/R^2$, we get $v=\sqrt{gR/2}$. The option $\sqrt{2gR}$ is the escape speed at the surface, $\sqrt{gR}$ ignores the extra distance, and $\frac{\sqrt{gR}}{2}$ misplaces the factor of 2 outside the square root.

3. Question 3

   A 2.0 kg stone is released from rest at the top of a 10 m building on Earth (g = $9.8\mathrm{m}{\mathrm{s}}^{-2}$). Simultaneously, a 2.0 kg satellite moves in a circular orbit 300 km above Earth’s surface (Earth radius $R=6.37\times 10^6\mathrm{m}$). Which statement correctly compares the magnitude of the gravitational force on the stone when it reaches the ground with that on the satellite in orbit?
   • AThe satellite experiences a larger force than the stone.
   • BThe forces cannot be compared without knowing the satellite's speed.
   • CThe stone experiences a larger force than the satellite.
   • DBoth experience equal forces.
   Show answer and explanation

   Answer: C) The stone experiences a larger force than the satellite.

   The gravitational force between Earth and any object of mass $m$ is given by Newton's law of universal gravitation: $F=\frac{GMm}{r^2}$, where $r$ is the distance from the center of the Earth to the object. For the stone on the ground, this distance is equal to the Earth's radius, $r_{\text{stone}}=R=6.37\times 10^6\mathrm{m}$. For the satellite in orbit, the distance is $r_{\text{satellite}}=R+h=6.37\times 10^6\mathrm{m}+300\times 10^3\mathrm{m}=6.67\times 10^6\mathrm{m}$. Since both objects have the same mass ($2.0\mathrm{kg}$) and $r_{\text{satellite}}>r_{\text{stone}}$, the gravitational force on the satellite is weaker than that on the stone. Thus, the stone experiences a larger gravitational force ($19.6\mathrm{N}$) than the satellite ($\approx 17.9\mathrm{N}$).

4. Question 4

   A satellite of mass $m$ orbits Earth in a stable circular orbit of radius $r$, where the gravitational pull supplies the necessary centripetal force. The satellite's engines perform a maneuver that moves it into a new stable circular orbit of radius $2r$. What is the satellite's new kinetic energy in terms of its initial kinetic energy $K$?
   • A$\frac{K}{2}$
   • B$2K$
   • C$\frac{K}{4}$
   • D$4K$
   Show answer and explanation

   Answer: A) $\frac{K}{2}$

   In a stable circular orbit the kinetic energy is $K=\frac{GMm}{2r}$. Doubling the orbital radius halves the kinetic energy, giving $K^{\prime }=\frac{K}{2}$. The option $\frac{K}{4}$ assumes an inverse‑square dependence, $2K$ assumes kinetic energy grows with radius, and $4K$ assumes a square‑law increase, all of which are incorrect.

5. Question 5

   Which force provides the centripetal acceleration required to keep planets in orbit around the Sun?
   • ANuclear force
   • BGravitational force
   • CElectrostatic force
   • DMagnetic force
   Show answer and explanation

   Answer: B) Gravitational force

   The gravitational attraction between the Sun and a planet acts toward the Sun and provides the necessary centripetal force to keep the planet in its nearly circular orbit. Electrostatic force acts between electric charges and is irrelevant because planets are overall neutral. Magnetic force requires moving charges or magnetic dipoles and does not dominate planetary motion. Nuclear force operates only over sub‑atomic distances inside nuclei and cannot affect planetary scales.

6. Question 6

   If the distance between two spherical masses is reduced to half of its original value, how does the magnitude of the gravitational force between them change?
   • AIt increases to twice the original force.
   • BIt decreases to half of the original force.
   • CIt decreases to one-fourth of the original force.
   • DIt increases to four times the original force.
   Show answer and explanation

   Answer: D) It increases to four times the original force.

   The gravitational force varies inversely with the square of the separation distance. Halving the distance reduces the denominator by a factor of $4$, so the force increases by the same factor, giving four times the original force. Options that suggest a decrease or only a linear increase misapply the inverse‑square relationship.

7. Question 7

   Why do astronauts inside an orbiting space station experience weightlessness even though Earth's gravity still acts on them?
   • AThe gravitational force exerted by the Moon balances Earth's pull.
   • BThe station's speed completely cancels out Earth's gravitational force.
   • CThe atmosphere at that altitude shields them from gravitational force.
   • DThey and the station are in continuous free fall toward Earth.
   Show answer and explanation

   Answer: D) They and the station are in continuous free fall toward Earth.

   In orbit the station and its occupants are falling toward Earth under gravity, but their forward velocity creates a continuous free‑fall trajectory, so no contact force is felt. This explains weightlessness. The other options incorrectly claim that speed cancels gravity, that the atmosphere shields gravity, or that the Moon balances Earth’s pull, all of which are misconceptions.

8. Question 8

   The acceleration due to gravity on the surface of a planet is $g$. What is the acceleration due to gravity at an altitude equal to twice the radius of the planet above its surface?
   • A$\frac{g}{9}$
   • B$\frac{g}{3}$
   • C$\frac{g}{2}$
   • D$\frac{g}{4}$
   Show answer and explanation

   Answer: A) $\frac{g}{9}$

   Gravitational acceleration follows the inverse‑square law: $g^{\prime }=g{\left(\frac{R}{r}\right)}^2$. The altitude is $2R$, so the distance from the centre is $3R$, giving $g^{\prime }=g/(3^2)=g/9$. The other options assume a linear decrease or ignore the planet’s radius, which is why they are incorrect.

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• Kinematics (5)
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