Question

The period of a satellite in a circular orbit of radius 12000 km around a planet is 3 hours. Obtain the period of a satellite in a circular orbit of radius 48000 km around the same planet.

• A. 6 hours
• B. 12 hours
• C. 24 hours
• D. 36 hours

Hint:

For problems involving Kepler's Third Law, always set up the ratio \( (T_2/T_1)^2 = (r_2/r_1)^3 \). It simplifies the calculation and you don't need to know the mass of the central planet or the gravitational constant.

Answer 1

The Correct Option is C

Step 1: State Kepler's Third Law of planetary motion. For objects orbiting the same central body, the square of the orbital period (\(T\)) is directly proportional to the cube of the semi-major axis of its orbit. For a circular orbit, this is the radius (\(r\)). \[ \frac{T^2}{r^3} = \text{constant} \]

Step 2: Set up a ratio for the two satellites. Let \(T_1, r_1\) be the period and radius for the first satellite, and \(T_2, r_2\) for the second. \[ \frac{T_1^2}{r_1^3} = \frac{T_2^2}{r_2^3} \]

Step 3: Rearrange the formula to solve for the unknown period, \(T_2\). \[ T_2^2 = T_1^2 \left( \frac{r_2}{r_1} \right)^3 \] \[ T_2 = T_1 \left( \frac{r_2}{r_1} \right)^{3/2} \]

Step 4: Substitute the given values and calculate \(T_2\). - \( T_1 = 3 \) hours - \( r_1 = 12000 \) km - \( r_2 = 48000 \) km The ratio of the radii is \( \frac{r_2}{r_1} = \frac{48000}{12000} = 4 \). \[ T_2 = 3 \times (4)^{3/2} = 3 \times (\sqrt{4})^3 = 3 \times (2)^3 = 3 \times 8 = 24 \text{ hours} \]