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CodeVID-P11-07-CH-01
Gravitation — Full Chapter Test
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- This is a full-length test covering the whole chapter — every topic is included.
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks. Show all working for Sections B, C and D.
Section A — Multiple Choice Questions
6 × 1 = 6 marks
1.
Newton's law of gravitation gives the force as:
- A.$F=\frac{Gm_1m_2}{r}$
- B.$F=\frac{Gm_1m_2}{r^2}$
- C.$F=Gm_1m_2r^2$
- D.$F=\frac{m_1m_2}{Gr^2}$
2.
The value of $g$ at the poles compared with the equator is:
- A.smaller
- B.larger
- C.equal
- D.zero
3.
Gravitational potential energy at infinity is taken as:
- A.maximum positive
- B.zero
- C.minimum negative
- D.undefined
4.
Escape velocity from the Earth is about:
- A.$7.9\ \text{km/s}$
- B.$11.2\ \text{km/s}$
- C.$9.8\ \text{km/s}$
- D.$3\times10^8\ \text{m/s}$
5.
The total energy of an orbiting satellite is:
- A.positive
- B.negative
- C.zero
- D.infinite
6.
Kepler's third law is:
- A.$T\propto r$
- B.$T^2\propto r^3$
- C.$T^3\propto r^2$
- D.$T^2\propto r$
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
7.
State Newton's law of gravitation.
8.
Define escape velocity and give its value for Earth.
9.
Why is $g$ zero at the centre of the Earth?
10.
State Kepler's law of areas.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
11.
Derive the expression for the variation of $g$ with depth $d$.
12.
A satellite orbits at radius $2R$. Find its orbital velocity in terms of $g$ and $R$.
Section D — Long Answer (5 marks)
2 × 5 = 10 marks
13.
Derive the expression for escape velocity from the surface of a planet and show that $v_e=\sqrt{2}\,v_o$, where $v_o$ is the orbital velocity near the surface.
14.
State Kepler's three laws of planetary motion and obtain the relation $T^2\propto r^3$ for a circular orbit.
Answer Key
Section A — Multiple Choice Questions
- (B) $F=\frac{Gm_1m_2}{r^2}$
- (B) larger
- (B) zero
- (B) $11.2\ \text{km/s}$
- (B) negative
- (B) $T^2\propto r^3$
Section B — Short Answer (2 marks)
- Two point masses attract with $F=\frac{Gm_1m_2}{r^2}$ along the line joining them.
- Minimum speed to escape gravity; $v_e=\sqrt{2gR}\approx11.2$ km/s for Earth.
- $g_d=g(1-d/R)$, which is zero when $d=R$ (the centre); the surrounding shell exerts no net pull there.
- The line joining a planet to the Sun sweeps out equal areas in equal intervals of time.
Section C — Short Answer (3 marks)
- $g_d=g\left(1-\frac{d}{R}\right)$, since only the mass within radius $(R-d)$ attracts the body.
- $v_o=\sqrt{\frac{GM}{2R}}=\sqrt{\frac{gR}{2}}$.
Section D — Long Answer (5 marks)
- From energy conservation $\frac{1}{2}mv_e^2=\frac{GMm}{R}$, so $v_e=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}$. Since $v_o=\sqrt{gR}$ near the surface, $v_e=\sqrt{2}\,v_o$.
- Laws of orbits (ellipse), areas (equal areas in equal times), and periods ($T^2\propto r^3$). For a circular orbit $\frac{GMm}{r^2}=\frac{mv^2}{r}$ and $T=\frac{2\pi r}{v}$ give $T^2=\frac{4\pi^2}{GM}r^3$, hence $T^2\propto r^3$.
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