The time period of a geostationary satellite is $24 ext{ hr}$ at a height $6R_E$ ($R_E$ is the radius of the

Question

The time period of a geostationary satellite is $24 	ext{ hr}$ at a height $6R_E$ ($R_E$ is the radius of the Earth) from the surface of the earth. The time period of another satellite whose height is $2.5R_E$ from the surface will be:

Accepted Answer

1. **Kepler's Third Law:** The square of the time period of revolution of a planet (or satellite) is proportional to the cube of the semi-major axis (radius for circular orbit) of its orbit. [Eq. 7.7, Eq. 7.38]
   $$ T^2 \propto r^3 \implies \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3} $$
2. **Determine Orbital Radii:** The distance $r$ is measured from the center of the Earth. Given $R_E$ is the Earth's radius.
   - Satellite 1 (Geostationary): Height $h_1 = 6R_E$. Orbit radius $r_1 = R_E + h_1 = R_E + 6R_E = 7R_E$.
   - Satellite 2: Height $h_2 = 2.5R_E$. Orbit radius $r_2 = R_E + h_2 = R_E + 2.5R_E = 3.5R_E$.
3. **Calculate Ratio:**
   $$ \frac{T_2}{T_1} = \left( \frac{r_2}{r_1} ight)^{3/2} = \left( \frac{3.5R_E}{7R_E} ight)^{3/2} = \left( \frac{1}{2} ight)^{3/2} $$
   $$ \frac{T_2}{24} = \frac{1}{2\sqrt{2}} $$
4. **Solve for $T_2$:**
   $$ T_2 = \frac{24}{2\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2} 	ext{ hr} $$

Step-by-Step Solution

Practice All PHYSICS Questions