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CodeVID-P11-07-PEE-01
Gravitational PE & Escape Velocity — Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- 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. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
The gravitational potential energy is taken as zero at:
- A.the centre of the Earth
- B.the surface of the Earth
- C.infinity
- D.a height equal to $R$
2.
The escape velocity from Earth is about:
- A.$8\ \text{km/s}$
- B.$11.2\ \text{km/s}$
- C.$22.4\ \text{km/s}$
- D.$3\times10^8\ \text{m/s}$
3.
Gravitational potential $V$ at distance $r$ from mass $M$ equals:
- A.$-\frac{GM}{r}$
- B.$-\frac{GM}{r^2}$
- C.$\frac{GM}{r}$
- D.$-\frac{GMm}{r}$
4.
If the KE given to a body equals the magnitude of its surface PE, the body will:
- A.fall back
- B.orbit at the surface
- C.just escape to infinity
- D.stay at rest
5.
The relation between escape velocity and orbital velocity near the surface is:
- A.$v_e=v_o$
- B.$v_e=2v_o$
- C.$v_e=\sqrt{2}\,v_o$
- D.$v_e=\frac{v_o}{\sqrt{2}}$
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
Why is gravitational potential energy negative?
7.
Define escape velocity and write its formula.
8.
On which factors does escape velocity depend?
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Derive $v_e=\sqrt{2gR}$ from energy conservation.
10.
Find the escape velocity from a planet with $g=4.9\ \text{m/s}^2$ and $R=3.2\times10^6\ \text{m}$.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Explain gravitational potential energy and gravitational potential. Derive expressions for both and explain the significance of the negative sign, then use them to obtain the escape velocity.
Answer Key
Section A — Multiple Choice Questions
- (C) infinity
- (B) $11.2\ \text{km/s}$
- (A) $-\frac{GM}{r}$
- (C) just escape to infinity
- (C) $v_e=\sqrt{2}\,v_o$
Section B — Short Answer (2 marks)
- Because PE is taken as zero at infinity and the attractive field does positive work as masses approach, so $U=-\frac{GMm}{r}<0$.
- Minimum speed to escape a planet's gravity; $v_e=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}$.
- Only on the planet's mass and radius (equivalently $g$ and $R$); not on the body's mass or launch direction.
Section C — Short Answer (3 marks)
- Set $\frac{1}{2}mv_e^2=\frac{GMm}{R}$, so $v_e=\sqrt{\frac{2GM}{R}}$; using $GM=gR^2$ gives $v_e=\sqrt{2gR}$.
- $v_e=\sqrt{2\times4.9\times3.2\times10^6}=\sqrt{3.136\times10^7}\approx5.6\ \text{km/s}$.
Section D — Long Answer (5 marks)
- $U=-\frac{GMm}{r}$ and $V=-\frac{GM}{r}$, both zero at infinity (negative because the field is attractive). Equating $\frac{1}{2}mv_e^2$ to $|U|=\frac{GMm}{R}$ gives $v_e=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}\approx11.2$ km/s for Earth.
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