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Gravitation — NEET & JEE MCQs with answers
This chapter traces the development of the law of universal gravitation from Kepler's empirical laws and Newton's insight that the same inverse-square force governs both terrestrial gravity and planetary motion. It covers gravitational potential energy, escape speed, satellite orbits, and the energy of orbiting satellites.
Topics in this chapter
- 7.1 Introduction
- 7.2 Kepler's Laws
- 7.3 Universal Law of Gravitation
- 7.4 The Gravitational Constant
- 7.5 Acceleration Due to Gravity of the Earth
- 7.6 Acceleration Due to Gravity Below and Above the Surface of Earth
- 7.7 Gravitational Potential Energy
- 7.8 Escape Speed
- 7.9 Earth Satellites
- 7.10 Energy of an Orbiting Satellite
Free practice questions
A clock based on a pendulum runs correctly at the surface. Taken to a mine at depth $d = R_E/100$, the value of g there (using $g(1-d/R_E)$) becomes:
- A. $0.99\,g$
- B. $0.98\,g$
- C. $1.01\,g$
- D. $0.90\,g$
Show answer
Answer: A — g(d) = g(1 − d/R) = g(1 − 1/100) = 0.99 g. So g decreases by 1% to 0.99g.
NCERT Class 11 Physics · Ch 7 · 7.6 Acceleration Due to Gravity Below and Above the Surface of Earth
If the mass of the Earth were doubled while its radius remained unchanged, the surface acceleration due to gravity would become:
- A. $4g$
- B. $g/2$
- C. $2g$
- D. $g$
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Answer: C — $g = GM_E/R_E^2 \propto M_E$ at fixed $R_E$. Doubling $M_E$ doubles $g$, so $g' = 2g$.
NCERT Class 11 Physics · Ch 7 · 7.5 Acceleration Due to Gravity of the Earth
According to the text, Tycho Brahe recorded his observations of the planets using:
- A. A pendulum clock
- B. A telescope
- C. A microscope
- D. The naked eye
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Answer: D — The text specifies that Tycho Brahe spent his entire lifetime recording observations of the planets with the naked eye.
NCERT Class 11 Physics · Ch 7 · 7.1 Introduction
Equating the gravitational force to the required centripetal force for a satellite, the orbital speed V is given by V² =
- A. G M_E/(R_E + h)³
- B. G M_E (R_E + h)
- C. G M_E/(R_E + h)
- D. G M_E/(R_E + h)²
Show answer
Answer: C — NCERT Eq. (7.35): equating mV²/(R_E+h) = GmM_E/(R_E+h)² and cancelling m gives V² = G M_E/(R_E + h).
NCERT Class 11 Physics · Ch 7 · 7.9 Earth Satellites
At which point in its elliptical orbit does a planet have the greatest orbital speed?
- A. Perihelion (closest to the Sun)
- B. Speed is the same everywhere
- C. At the ends of the minor axis
- D. Aphelion (farthest from the Sun)
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Answer: A — By angular momentum conservation $rv$ = constant, so speed is greatest where $r$ is smallest — at the perihelion (closest point to the Sun).
NCERT Class 11 Physics · Ch 7 · 7.2 Kepler's Laws
The gravitational potential at a point due to a mass $M = 2\times10^{20}\,\text{kg}$ at distance $r = 1\times10^{7}\,\text{m}$ is (take $G=6.67\times10^{-11}$):
- A. $-1.33\times10^{3}\,\text{J kg}^{-1}$
- B. $-1.33\times10^{4}\,\text{J kg}^{-1}$
- C. $+1.33\times10^{3}\,\text{J kg}^{-1}$
- D. $-1.33\times10^{2}\,\text{J kg}^{-1}$
Show answer
Answer: A — $V = -\dfrac{GM}{r} = -\dfrac{6.67\times10^{-11}\times2\times10^{20}}{1\times10^{7}} = -\dfrac{1.334\times10^{10}}{1\times10^{7}} = -1.33\times10^{3}\,\text{J kg}^{-1}$. Section 7.7.
NCERT Class 11 Physics · Ch 7 · 7.7 Gravitational Potential Energy
Taking the potential energy at infinity as zero (W_∞ = 0), the total energy of an object that just reaches infinity with zero speed is:
- A. Negative
- B. Positive
- C. Zero
- D. Infinite
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Answer: C — At infinity, E(∞) = (1/2)mV_f^2 + W_∞. With W_∞ = 0 and V_f = 0 for the minimum escape case, the total energy is zero.
NCERT Class 11 Physics · Ch 7 · 7.8 Escape Speed
The potential energy of a satellite of mass $m$ at a distance $(R+h)$ from the centre of Earth (mass $M$), taking PE at infinity as zero, is:
- A. $-\dfrac{GMm}{(R+h)}$
- B. $-\dfrac{GMm}{2(R+h)}$
- C. $\dfrac{GMm}{(R+h)^2}$
- D. $+\dfrac{GMm}{(R+h)}$
Show answer
Answer: A — From Eq. (7.41), $PE = -\frac{GMm}{(R+h)}$. It is negative because PE at infinity is taken as zero and the satellite is bound.
NCERT Class 11 Physics · Ch 7 · 7.10 Energy of an Orbiting Satellite
In the vector form of Newton's gravitation law, the force on $m_2$ due to $m_1$ is $\mathbf{F} = -\dfrac{Gm_1 m_2}{r^2}\hat{\mathbf{r}}$, where $\hat{\mathbf{r}}$ is the unit vector from $m_1$ to $m_2$. The negative sign indicates that the force is:
- A. attractive (directed from $m_2$ toward $m_1$)
- B. zero
- C. repulsive (directed away from $m_1$)
- D. perpendicular to the line joining the masses
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Answer: A — The negative sign shows the force on $m_2$ points opposite to $\hat{\mathbf{r}}$ (from $m_1$ to $m_2$), i.e. it points back toward $m_1$ — confirming gravitation is attractive.
NCERT Class 11 Physics · Ch 7 · 7.3 Universal Law of Gravitation
The gravitational constant G is best described as:
- A. a universal constant determined experimentally
- B. a quantity that varies from place to place on the earth
- C. a constant that depends on the masses involved
- D. the acceleration due to gravity
Show answer
Answer: A — NCERT calls G 'the gravitational constant... entering the Universal law of gravitation' whose value is determined experimentally. It is a universal constant with the value 6.67×10⁻¹¹ N m²/kg².
NCERT Class 11 Physics · Ch 7 · 7.4 The Gravitational Constant
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