Question

Two satellites A and B are revolving around the earth in orbits of heights \(1.25R_E\) and \(19.25R_E\) from the surface of earth respectively, where \(R_E\) is the radius of the earth. The ratio of the orbital speeds of the satellites A and B is

• A. 5:1
• B. 4:1
• C. 9:1
• D. 3:1

Hint:

Orbital speed of a satellite around a central body of mass M at an orbital radius \(r\) (distance from centre of M) is \( v = \sqrt{\frac{GM}{r}} \). If height from surface is \(h\) and radius of central body is \(R\), then \( r = R+h \). The ratio of speeds \( v_1/v_2 = \sqrt{r_2/r_1} \).

Answer 1

The Correct Option is D

The orbital speed \(v\) of a satellite at a distance \(r\) from the centre of the Earth is given by \( v = \sqrt{\frac{GM}{r}} \), where G is the gravitational constant and M is the mass of the Earth.
The distance \(r\) is measured from the centre of the Earth.
Given heights are from the surface of the Earth.
For satellite A: Height \( h_A = 1.
25R_E \).
Orbital radius \( r_A = R_E + h_A = R_E + 1.
25R_E = 2.
25R_E \).
Orbital speed \( v_A = \sqrt{\frac{GM}{r_A}} = \sqrt{\frac{GM}{2.
25R_E}} \).
For satellite B: Height \( h_B = 19.
25R_E \).
Orbital radius \( r_B = R_E + h_B = R_E + 19.
25R_E = 20.
25R_E \).
Orbital speed \( v_B = \sqrt{\frac{GM}{r_B}} = \sqrt{\frac{GM}{20.
25R_E}} \).
Ratio of orbital speeds \( \frac{v_A}{v_B} \): \[ \frac{v_A}{v_B} = \frac{\sqrt{\frac{GM}{2.
25R_E}}}}{\sqrt{\frac{GM}{20.
25R_E}}}} = \sqrt{\frac{GM}{2.
25R_E} \cdot \frac{20.
25R_E}{GM}} = \sqrt{\frac{20.
25R_E}{2.
25R_E}} \] \[ \frac{v_A}{v_B} = \sqrt{\frac{20.
25}{2.
25}} \] To simplify the fraction: \( \frac{20.
25}{2.
25} = \frac{2025}{225} \).
\( 225 \times 10 = 2250 \).
\( 225 \times 9 = 2250 - 225 = 2025 \).
So, \( \frac{2025}{225} = 9 \).
\[ \frac{v_A}{v_B} = \sqrt{9} = 3 \] The ratio \( v_A : v_B = 3:1 \).
This matches option (4).