Vidaara.orgClass 11 · Physics
CodeVID-P11-07-SAT-01
Satellites & Kepler's Laws — 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 orbital velocity of a satellite is independent of:
- A.the orbit radius
- B.the mass of the planet
- C.the mass of the satellite
- D.the value of $G$
2.
The total energy of an orbiting satellite is:
- A.$+\frac{GMm}{2r}$
- B.$-\frac{GMm}{2r}$
- C.$-\frac{GMm}{r}$
- D.zero
3.
Kepler's law of areas is a consequence of the conservation of:
- A.linear momentum
- B.energy
- C.angular momentum
- D.charge
4.
The height of a geostationary satellite above the Earth's surface is about:
- A.$3600\ \text{km}$
- B.$36{,}000\ \text{km}$
- C.$360\ \text{km}$
- D.$3.6\times10^8\ \text{km}$
5.
For orbits very close to the Earth's surface, the orbital velocity is about:
- A.$5.6\ \text{km/s}$
- B.$7.9\ \text{km/s}$
- C.$11.2\ \text{km/s}$
- D.$3\times10^5\ \text{km/s}$
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
Define orbital velocity and write its formula.
7.
Why is the total energy of a satellite negative?
8.
State Kepler's third law and its mathematical form.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Derive the expression for the orbital velocity of a satellite.
10.
A planet's orbital radius is 9 times Earth's. Find its period in Earth years.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
State Kepler's three laws of planetary motion and explain the physical meaning of each, noting which conservation principle the second law expresses.
Answer Key
Section A — Multiple Choice Questions
- (C) the mass of the satellite
- (B) $-\frac{GMm}{2r}$
- (C) angular momentum
- (B) $36{,}000\ \text{km}$
- (B) $7.9\ \text{km/s}$
Section B — Short Answer (2 marks)
- Speed needed to stay in a circular orbit: $v_o=\sqrt{\frac{GM}{r}}$.
- Because $E=KE+PE=\frac{GMm}{2r}-\frac{GMm}{r}=-\frac{GMm}{2r}<0$, showing it is gravitationally bound.
- $T^2\propto r^3$, i.e. $\frac{T^2}{r^3}$ is the same constant for all planets around the Sun.
Section C — Short Answer (3 marks)
- Set $\frac{GMm}{r^2}=\frac{mv_o^2}{r}$, giving $v_o=\sqrt{\frac{GM}{r}}$; near the surface $v_o=\sqrt{gR}$.
- $T^2\propto r^3\Rightarrow T=9^{3/2}=27$ Earth years.
Section D — Long Answer (5 marks)
- First: planets move in ellipses with the Sun at one focus. Second: the radius vector sweeps equal areas in equal times (a planet moves faster near the Sun) — this expresses conservation of angular momentum. Third: $T^2\propto r^3$ for all planets around the Sun.
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