Vidaara.orgClass 9 · Physics
CodeVID-P9-03-ULG-01
Universal Law of Gravitation & g — 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. Take $G = 6.67 \times 10^{-11}\ \text{N m}^2/\text{kg}^2$ and $g = 9.8\ \text{m/s}^2$ unless stated otherwise.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
In $F = \dfrac{G m_1 m_2}{r^2}$, doubling $r$ makes $F$:
- A.double
- B.half
- C.one-quarter
- D.four times
2.
The SI unit of $G$ is:
- A.$\text{N m}^2/\text{kg}^2$
- B.$\text{m/s}^2$
- C.$\text{N/kg}$
- D.$\text{kg m}/\text{s}^2$
3.
$g$ at Earth's surface is about:
- A.$0.98\ \text{m/s}^2$
- B.$9.8\ \text{m/s}^2$
- C.$98\ \text{m/s}^2$
- D.$980\ \text{m/s}^2$
4.
The formula for $g$ is:
- A.$g = \dfrac{GM}{R}$
- B.$g = \dfrac{GM}{R^2}$
- C.$g = \dfrac{GMR^2}{1}$
- D.$g = \dfrac{R^2}{GM}$
5.
Gravitational force between two bodies is always:
- A.repulsive
- B.attractive
- C.zero
- D.sometimes both
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
State the universal law of gravitation.
7.
Why do we not feel the gravitational pull between two people standing nearby?
8.
Name two factors on which the gravitational force between two objects depends.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Two masses of $10\ \text{kg}$ each are $1\ \text{m}$ apart. Find the force between them.
10.
Derive $g = \dfrac{GM}{R^2}$ starting from $F = mg$.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Explain how $g$ varies with altitude, depth and latitude, giving the reason in each case.
Answer Key
Section A — Multiple Choice Questions
- (C) one-quarter
- (A) $\text{N m}^2/\text{kg}^2$
- (B) $9.8\ \text{m/s}^2$
- (B) $g = \dfrac{GM}{R^2}$
- (B) attractive
Section B — Short Answer (2 marks)
- Every object attracts every other with a force $F = \dfrac{G m_1 m_2}{r^2}$, proportional to the product of masses and inversely proportional to the square of the distance.
- Because $G$ is extremely small and human masses are small, the force is negligibly tiny.
- The product of their masses and the square of the distance between their centres.
Section C — Short Answer (3 marks)
- $F = \dfrac{(6.67\times10^{-11})(10)(10)}{1^2} = 6.67 \times 10^{-9}\ \text{N}$.
- At the surface $F = \dfrac{GMm}{R^2}$ and $F = mg$; equate and cancel $m$ to get $g = \dfrac{GM}{R^2}$.
Section D — Long Answer (5 marks)
- Altitude: $g$ decreases as $r$ increases ($g\propto1/r^2$). Depth: $g$ decreases because only the mass below contributes, reaching 0 at the centre. Latitude: $g$ is greatest at the poles (smaller $R$) and least at the equator (larger $R$ plus rotation effect).
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