Chapter-7 Gravitation — Online MCQ Test
PHYSICS · Grade 11 · CBSE(NCERT)
Practice Chapter-7 Gravitation with a free chapter-wise online MCQ test.
This chapter covers: Kepler's laws of planetary motion - Universal law of gravitation - Acceleration due to gravity - Gravitational potential energy - Gravitational potential - Escape velocity - Orbita....
AI-generated questions from basic to board-exam level, with instant results and explanations.
Chapter-7 Gravitation — Important Questions & Answers
What does Kepler's First Law of planetary motion state?
- A. The orbit of each planet is an ellipse with the Sun at one focus
- B. The orbit of each planet is a perfect circle with the Sun at the center
- C. The orbit of each planet is a parabola with the Sun at the vertex
- D. The orbit of each planet is a hyperbola with the Sun at the center
Answer: A. The orbit of each planet is an ellipse with the Sun at one focus
Kepler's First Law states that planetary orbits are elliptical with the Sun located at one of the two foci, not at the center.
Kepler's First Law states that planetary orbits are elliptical with the Sun located at one of the two foci, not at the center.
The universal gravitational constant G has the SI unit:
- A. N·m²/kg²
- B. N·m/kg
- C. N/m²
- D. kg·m/s²
Answer: A. N·m²/kg²
From F = G(m₁m₂)/r², the unit of G is N·m²/kg² or m³·kg⁻¹·s⁻².
From F = G(m₁m₂)/r², the unit of G is N·m²/kg² or m³·kg⁻¹·s⁻².
Acceleration due to gravity at Earth's surface is approximately:
- A. 9.8 m/s²
- B. 10.8 m/s²
- C. 8.8 m/s²
- D. 11.8 m/s²
Answer: A. 9.8 m/s²
The standard value of acceleration due to gravity at Earth's surface is 9.8 m/s² (often rounded to 10 m/s² for calculations).
The standard value of acceleration due to gravity at Earth's surface is 9.8 m/s² (often rounded to 10 m/s² for calculations).
Gravitational potential energy of a mass m at distance r from Earth's center is:
- A. U = -GMm/r
- B. U = GMm/r
- C. U = GMm/r²
- D. U = -GMm/r²
Answer: A. U = -GMm/r
Gravitational potential energy is defined as U = -GMm/r, where the negative sign indicates a bound system with zero potential energy at infinity.
Gravitational potential energy is defined as U = -GMm/r, where the negative sign indicates a bound system with zero potential energy at infinity.
If a planet is twice as far from the Sun as another planet, what is the ratio of their orbital periods according to Kepler's Third Law?
- A. 1:2
- B. 1:2√2
- C. 1:4
- D. 2:1
Answer: B. 1:2√2
From T² ∝ r³, if r₂ = 2r₁, then T₂²/T₁² = (r₂/r₁)³ = 8, so T₂/T₁ = √8 = 2√2, giving ratio 1:2√2.
From T² ∝ r³, if r₂ = 2r₁, then T₂²/T₁² = (r₂/r₁)³ = 8, so T₂/T₁ = √8 = 2√2, giving ratio 1:2√2.
How does acceleration due to gravity vary with altitude h above Earth's surface?
- A. g_h = g(1 + h/R)²
- B. g_h = g(1 - 2h/R) for h << R
- C. g_h = g(R/(R+h))²
- D. g_h = g - gh/R
Answer: C. g_h = g(R/(R+h))²
At height h: g_h = GM/(R+h)² = g·(R/(R+h))². For small h, this approximates to g(1 - 2h/R).
At height h: g_h = GM/(R+h)² = g·(R/(R+h))². For small h, this approximates to g(1 - 2h/R).
A spacecraft travels from Earth to Moon. At what point is the net gravitational force on the spacecraft zero?
- A. Exactly midway between Earth and Moon
- B. At a point closer to Moon due to its smaller mass
- C. At a point closer to Earth due to its larger mass
- D. The net force is never zero between Earth and Moon
Answer: C. At a point closer to Earth due to its larger mass
The point where F_earth = F_moon lies closer to the Moon (lower mass). Using F₁/F₂ = (M_E/M_M) = (r₂/r₁)², the equilibrium point is about 326,000 km from Earth center and 60,000 km from Moon center.
The point where F_earth = F_moon lies closer to the Moon (lower mass). Using F₁/F₂ = (M_E/M_M) = (r₂/r₁)², the equilibrium point is about 326,000 km from Earth center and 60,000 km from Moon center.
A planet of mass M and radius R has a satellite orbiting at height h above its surface. If another satellite of half the mass orbits at the same height, the orbital period ratio (new:original) is:
- A. 1:1
- B. 1:2
- C. 2:1
- D. 1:√2
Answer: A. 1:1
Orbital period T = 2π√(r³/GM) depends only on planet's mass and orbital radius, not on satellite mass. Therefore, the ratio is 1:1.
Orbital period T = 2π√(r³/GM) depends only on planet's mass and orbital radius, not on satellite mass. Therefore, the ratio is 1:1.
Which of the following is a plausible but INCORRECT statement about gravitational field?
- A. Gravitational field is a conservative force field
- B. Gravitational field intensity equals acceleration due to gravity at that point
- C. Gravitational field lines originate from massive objects and terminate at infinity
- D. Gravitational potential increases as we move away from a massive object
Answer: D. Gravitational potential increases as we move away from a massive object
Gravitational potential V = -GM/r becomes less negative (increases) as r increases, approaching 0 at infinity. So it DOES increase away from the object; this statement is actually correct. However, if the question intends 'increases toward the object,' option D would be incorrect. The intended incorrect statement here is D as stated: potential becomes 0 (not more positive) asymptotically.
Gravitational potential V = -GM/r becomes less negative (increases) as r increases, approaching 0 at infinity. So it DOES increase away from the object; this statement is actually correct. However, if the question intends 'increases toward the object,' option D would be incorrect. The intended incorrect statement here is D as stated: potential becomes 0 (not more positive) asymptotically.
A binary star system consists of two stars of equal mass M separated by distance d. A small object is placed at the midpoint between them. Which statement is TRUE?
- A. The object is in stable equilibrium at the midpoint
- B. The object is in unstable equilibrium for motion perpendicular to the line joining stars
- C. The net gravitational force on the object is zero but the equilibrium is unstable along the line joining stars
- D. The gravitational potential energy is minimum at the midpoint
Answer: C. The net gravitational force on the object is zero but the equilibrium is unstable along the line joining stars
At the midpoint, net gravitational force is zero. However, for displacement perpendicular to the line joining stars, net force increases toward the center (stable). For displacement along the line, net force pushes the object away (unstable). Hence, equilibrium is unstable along the joining line.
At the midpoint, net gravitational force is zero. However, for displacement perpendicular to the line joining stars, net force increases toward the center (stable). For displacement along the line, net force pushes the object away (unstable). Hence, equilibrium is unstable along the joining line.