Gravitation Quiz - ICSE Class 11 Physics
Free practice quiz: Gravitation (ICSE Class 11 Physics). 51 multiple-choice questions with answers and explanations. Play online, no sign-up needed.
Play all 51 questions freeSample questions
A satellite is orbiting the Earth in a circular orbit of radius R. If the satellite's speed is V, what is the force of gravity experienced by the satellite?
- mV²/R
- mR²/V
- mV/R²
- mR/V²
Show answer
Answer: A. mV²/R
For a satellite in a circular orbit, the centripetal force required to maintain the orbit is provided by the force of gravity. The centripetal force is given by F = mV²/R, where m is the mass of the satellite, V is its orbital speed, and R is the radius of the orbit.
A planet has a mass twice that of Earth and a radius half that of Earth. What is the ratio of escape velocity from this planet to that from Earth?
- $\sqrt{2}$
- 2
- $2\sqrt{2}$
- 4
Show answer
Answer: C. $2\sqrt{2}$
The escape velocity from the surface of a planet is given by $v_e = \sqrt{2GM/R}$. Let the Earth's mass be $M_E$ and radius be $R_E$. The escape velocity from Earth is $v_{eE} = \sqrt{2GM_E/R_E}$. For the new planet, its mass is $M_P = 2M_E$ and its radius is $R_P = R_E/2$. The escape velocity from this planet is $v_{eP} = \sqrt{2GM_P/R_P} = \sqrt{2G(2M_E)/(R_E/2)} = \sqrt{2G(4M_E)/R_E} = \sqrt{4} \sqrt{2GM_E/R_E} = 2 v_{eE}$. Wait, let me recheck the calculation. $v_{eP} = \sqrt{2G(2M_E)/(R_E/2)} = \sqrt{2G \cdot 2M_E \cdot (2/R_E)} = \sqrt{8 GM_E/R_E} = \sqrt{4} \sqrt{2GM_E/R_E} = 2 imes \sqrt{2} v_{eE}$. So the ratio $v_{eP}/v_{eE} = 2\sqrt{2}$. My previous calculation was incorrect. The correct answer is $2\sqrt{2}$.
A mass m is orbiting the Earth in a circular orbit of radius R. If the mass of the Earth doubles but the radius of the orbit remains the same, how does the orbital velocity change?
- Increases by a factor of $\sqrt{2}$
- Decreases by a factor of $\sqrt{2}$
- Doubles
- Halves
Show answer
Answer: A. Increases by a factor of $\sqrt{2}$
For a satellite in a circular orbit, the orbital velocity v is given by $v = \sqrt{GM/R}$, where M is the mass of the planet and R is the radius of the orbit. If the mass of the Earth doubles (M becomes 2M) and the radius R remains the same, the new orbital velocity $v'$ will be $v' = \sqrt{G(2M)/R} = \sqrt{2} \sqrt{GM/R} = \sqrt{2} v$. Therefore, the orbital velocity increases by a factor of $\sqrt{2}$.
A satellite of mass m is in a circular orbit of radius r around a planet of mass M. What is the magnitude of the gravitational force acting on the satellite?
- $G M m / r$
- $G M m / r^2$
- $G m / r^2$
- $G M / r^2$
Show answer
Answer: B. $G M m / r^2$
Newton's law of universal gravitation states that the force of gravity between two objects of masses M and m separated by a distance r is given by $F = G \frac{M m}{r^2}$. In this case, M is the mass of the planet, m is the mass of the satellite, and r is the radius of the orbit (distance between their centers).
A meteoroid is falling towards the Earth. As it enters the Earth's atmosphere, its speed increases significantly due to air resistance and gravitational acceleration. What is the primary reason for its initial acceleration towards Earth?
- Centrifugal force
- Gravitational force
- Centripetal force
- Inertial force
Show answer
Answer: B. Gravitational force
The primary force causing the meteoroid to accelerate towards the Earth is the Earth's gravitational pull. Air resistance acts as a retarding force, opposing motion, and while it can affect the speed, the initial and sustained acceleration towards Earth is due to gravity.