Test Index

Gravitation Part 2

Section: Physics

© examsnet.com

Question : 10 of 110

Marks: +1, -0

Initially a satellite of 100 kg is in a circular orbit of radius

1.5 R_{E}

. This satellite can be moved to a circular orbit of radius

3 R_E

\alpha \times 10^{6} \text{J}

of energy The value of

\alpha

\_\_\_\_

(Take Radius of Earth

R_E = 6 \times 10^{6} \text{m}

g = 10 \text{m/s}^2

[21 Jan 2026 Shift 1]

500
1000
100
150

Solution:

Concept:

The total energy of a satellite in a circular orbit is

-\frac{GMm}{2r}

, where

r

is the orbital radius.

Explanation:

Given: satellite mass

m = 100 \text{ kg}

, initial radius

r_1 = 1.5R_E

, final radius

r_2 = 3R_E

,

R_E = 6 \times 10^6 \text{ m}

,

g = 10 \text{ m/s}^2

.

Using

g = \frac{GM}{R_E^2}

, we get

GM = g R_E^2

.

The total energy at radius

r

is

E = -\frac{GMm}{2r} = -\frac{mg R_E^2}{2r}

.

Energy supplied

\Delta E = E_2 - E_1 = \frac{mg R_E^2}{2} \left( \frac{1}{r_1} - \frac{1}{r_2} \right)

.

Substitute

r_1 = 1.5R_E

and

r_2 = 3R_E

:

\frac{1}{r_1} - \frac{1}{r_2} = \frac{1}{1.5R_E} - \frac{1}{3R_E} = \frac{2}{3R_E} - \frac{1}{3R_E} = \frac{1}{3R_E}

.

Thus

\Delta E = \frac{mgR_E^2}{2} \cdot \frac{1}{3R_E} = \frac{mgR_E}{6}

.

Plug in values:

m=100

,

g=10

,

R_E=6 \times 10^6

. Then

\Delta E = \frac{100 \times 10 \times 6 \times 10^6}{6} = 1000 \times 10^6 \text{ J} = \alpha \times 10^6 \text{ J}

.

Therefore

\alpha = 1000

.

Answer:

\alpha = 1000

, which corresponds to option (B).

© examsnet.com

Go to Question:

Prev Question

Next Question