Question

The time period of a 1500 kg satellite is equal to the time period of rotation of the earth. The altitude of the satellite is nearly

• A. \( 42,211 \) km
• B. \( 35,840 \) km
• C. \( 6,400 \) km
• D. \( 13,800 \) km

Hint:

A satellite with a time period equal to the Earth's rotation period is a geostationary satellite. Use Kepler's third law to relate the time period and the orbital radius. Remember to use consistent units (SI units are recommended). The distance from the Earth's center \( r \) includes the Earth's radius \( R_E \) and the altitude \( h \) of the satellite.

Answer 1

The Correct Option is B

A satellite whose time period of revolution is equal to the time period of rotation of the Earth (24 hours) is called a geostationary satellite.
A geostationary satellite appears to be stationary with respect to a point on the Earth's surface.
The time period of a satellite orbiting the Earth at a distance \( r \) from the center of the Earth is given by Kepler's third law: $$ T^2 = \frac{4\pi^2}{GM_E} r^3 $$ where \( G \) is the gravitational constant and \( M_E \) is the mass of the Earth.
The time period of rotation of the Earth is \( T = 24 \) hours \( = 24 \times 3600 \) seconds \( = 86400 \) s.
We need to find the altitude \( h \) of the satellite above the Earth's surface.
The distance from the center of the Earth is \( r = R_E + h \), where \( R_E \) is the radius of the Earth (approximately 6400 km).
Substituting the values into Kepler's third law: $$ (86400)^2 = \frac{4\pi^2}{(6.
674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2)(5.
972 \times 10^{24} \text{ kg})} r^3 $$ $$ (86400)^2 = \frac{4\pi^2}{3.
986 \times 10^{14}} r^3 $$ $$ 7.
465 \times 10^9 = 9.
95 \times 10^{-15} r^3 $$ $$ r^3 = \frac{7.
465 \times 10^9}{9.
95 \times 10^{-15}} = 7.
50 \times 10^{23} $$ $$ r = (7.
50 \times 10^{23})^{1/3} = 4.
216 \times 10^7 \text{ m} = 42,160 \text{ km} $$ The altitude \( h = r - R_E = 42,160 \text{ km} - 6,400 \text{ km} = 35,760 \text{ km} \).
This is approximately \( 35,840 \) km.