Class XI · Chapter 07
Gravitation — Exam Practice
Practice questions in CBSE, NEET, and JEE Main style covering Kepler’s laws, orbital velocity, escape speed, and satellite motion.
A note on these questions: these are original practice questions, written to match the exact style, format, and difficulty of CBSE Board, NEET, and JEE Main papers — not verbatim reproductions of specific past papers.
1CBSE Board Exam Style
Q1. Why is the value of acceleration due to gravity, g, zero at the centre of the Earth?
Answer
At the exact centre, the mass of Earth surrounding that point pulls equally in all directions, cancelling out completely by symmetry — leaving a net gravitational field of zero.
At the exact centre, the mass of Earth surrounding that point pulls equally in all directions, cancelling out completely by symmetry — leaving a net gravitational field of zero.
Derive the expression for the orbital velocity of a satellite moving in a circular orbit just above Earth’s surface.
Answer
Gravity supplies the centripetal force: GMm/R² = mv²/R.
Gravity supplies the centripetal force: GMm/R² = mv²/R.
Solving for v: v² = GM/R → v = √(GM/R). Since g = GM/R², this can also be written as v = √(gR).
Q3. (a) Derive the expression for escape velocity from a planet’s surface. (b) Find the escape velocity from a planet with mass twice Earth’s mass and the same radius as Earth. (Escape velocity of Earth = 11.2 km/s)
Answer
(a) Escape velocity is the minimum launch speed for which total mechanical energy is exactly zero at infinity: ½mve² − GMm/R = 0 → ve = √(2GM/R).
(a) Escape velocity is the minimum launch speed for which total mechanical energy is exactly zero at infinity: ½mve² − GMm/R = 0 → ve = √(2GM/R).
(b) Since ve ∝ √(M/R), and R is unchanged while M doubles: ve,planet = ve,Earth × √2 = 11.2 × 1.414 ≈ 15.8 km/s.
2NEET Style (Single-Correct MCQ)
Q4. Acceleration due to gravity, g, varies with:
Solution
g decreases with height (farther from centre), decreases with depth (less enclosed mass), and varies slightly with latitude (Earth’s equatorial bulge and rotation effects).
g decreases with height (farther from centre), decreases with depth (less enclosed mass), and varies slightly with latitude (Earth’s equatorial bulge and rotation effects).
Q5. The time period of a geostationary satellite is:
Solution
A geostationary satellite’s period exactly matches Earth’s rotation period, so it stays fixed above one point on the equator.
A geostationary satellite’s period exactly matches Earth’s rotation period, so it stays fixed above one point on the equator.
Q6. Gravitational potential energy of a two-mass system (taking zero at infinite separation) is always:
Solution
U = −GMm/r — negative at every finite separation, reflecting that gravity is attractive and work must be done to pull the masses apart to infinity.
U = −GMm/r — negative at every finite separation, reflecting that gravity is attractive and work must be done to pull the masses apart to infinity.
Q7. Kepler’s third law states that, for planets orbiting the same star:
Solution
The square of the orbital period is proportional to the cube of the semi-major axis — later derived directly from Newton’s law of gravitation.
The square of the orbital period is proportional to the cube of the semi-major axis — later derived directly from Newton’s law of gravitation.
3JEE Main Style
Q8. A satellite orbits at a height equal to Earth’s radius R above the surface. Express its orbital velocity in terms of v₀, the orbital velocity for a satellite very close to Earth’s surface.
Solution
Orbital radius r = R + R = 2R. v = √(GM/r) = √(GM/2R) = v₀/√2, since v₀ = √(GM/R).
Orbital radius r = R + R = 2R. v = √(GM/r) = √(GM/2R) = v₀/√2, since v₀ = √(GM/R).
The new orbital velocity is v₀/√2 ≈ 0.707v₀.
Q9. Astronauts orbiting Earth experience weightlessness because:
Solution
Gravity is only slightly weaker at typical orbital altitudes — the sensation of weightlessness comes from falling around Earth at the same rate as the spacecraft, not from an absence of gravity.
Gravity is only slightly weaker at typical orbital altitudes — the sensation of weightlessness comes from falling around Earth at the same rate as the spacecraft, not from an absence of gravity.
Q10. A satellite orbits Earth at a height of 2000 km above the surface. Estimate its orbital period. (R = 6400 km, g = 9.8 m/s²)
Solution
r = R + h = 6400 + 2000 = 8400 km = 8.4×10⁶ m. GM = gR² = 9.8×(6.4×10⁶)² ≈ 4.01×10¹⁴.
r = R + h = 6400 + 2000 = 8400 km = 8.4×10⁶ m. GM = gR² = 9.8×(6.4×10⁶)² ≈ 4.01×10¹⁴.
T = 2π√(r³/GM) = 2π√[(8.4×10⁶)³ / 4.01×10¹⁴] ≈ 2π×1216 ≈ 7640 s ≈ 127 minutes.
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