Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

The time period of a geostationary satellite is $24 \text{ hr}$ 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:

$6\sqrt{2} \text{ hr}$

$12\sqrt{2} \text{ hr}$

$\frac{24}{2.5} \text{ hr}$

$\frac{12}{2.5} \text{ hr}$

Step-by-Step Solution

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] $T^2 \propto r^3 \implies \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}$
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 $r_2 = R_E + h_2 = R_E + 2.5R_E = 3.5R_E$ .

Calculate Ratio: $\frac{T_2}{T_1} = \left( \frac{r_2}{r_1} \right)^{3/2} = \left( \frac{3.5R_E}{7R_E} \right)^{3/2} = \left( \frac{1}{2} \right)^{3/2}$ $\frac{T_2}{24} = \frac{1}{2\sqrt{2}}$
Solve for $T_2$ : $T_2 = \frac{24}{2\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2} \text{ hr}$

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