Question

The period of revolution of a communication satellite is 24 hours. The period of a satellite in an orbit at a distance three times that of the earth’s radius above its surface will be

• A. 4 days
• B. 16 days
• C. 24 hours
• D. 8 days

Hint:

For satellites, always apply Kepler’s third law using the orbital radius measured from the center of the Earth, not just the height above the surface.

Answer 1

The Correct Option is D

Step 1: Understanding the given data.
The period of revolution of a communication satellite is given as 24 hours. A communication satellite revolves very close to the Earth, so its orbital radius can be taken approximately equal to the radius of the Earth, i.e., \[ r_1 = R \] and \[ T_1 = 24 \text{ hours} \] The new satellite is at a height equal to three times the Earth's radius above the surface. Therefore, its orbital radius becomes: \[ r_2 = R + 3R = 4R \] Step 2: Applying Kepler’s third law.
According to Kepler’s third law of planetary motion: \[ T^2 \propto r^3 \] So, \[ \left(\frac{T_2}{T_1}\right)^2 = \left(\frac{r_2}{r_1}\right)^3 \] Substituting the values: \[ \left(\frac{T_2}{24}\right)^2 = \left(\frac{4R}{R}\right)^3 = 4^3 = 64 \] Step 3: Calculating the time period.
\[ \frac{T_2}{24} = \sqrt{64} = 8 \] \[ T_2 = 24 \times 8 = 192 \text{ hours} \] \[ 192 \text{ hours} = \frac{192}{24} = 8 \text{ days} \] Step 4: Final conclusion.
The period of revolution of the satellite at the given height is 8 days.