Question

The time period of revolution of a satellite close to the planet’s surface is 80 minutes. The time period of another satellite which is at a height of 3 times the radius of the planet from the surface is:

• A. \( 64 \) minutes
• B. \( 640 \) minutes
• C. \( 320 \) minutes
• D. \( 240 \) minutes

Hint:

Kepler’s third law states that the time period of a satellite increases as the cube root of the orbital radius. The farther the satellite, the longer its revolution time.

Answer 1

The Correct Option is B

Step 1: Use Kepler’s Third Law Kepler’s third law states that the square of the orbital period \( T \) is proportional to the cube of the semi
-major axis \( R \): \[ T^2 \propto R^3 \]
Step 2: Calculate the Ratio of Time Periods For a satellite near the planet’s surface: \[ T_1 = 80 \text{ minutes} \] The radius of the second satellite’s orbit is: \[ R_2 = R + 3R = 4R \] Using Kepler’s law: \[ \left( \frac{T_2}{T_1} \right)^2 = \left( \frac{R_2}{R_1} \right)^3 \] \[ \left( \frac{T_2}{80} \right)^2 = \left( \frac{4R}{R} \right)^3 = 4^3 = 64 \] \[ T_2 = 80 \times 8 = 640 \text{ minutes} \]