The time period of a satellite of earth is 5 hours . If the separation between the earth and the satellite is increased to 4 times the previous value, - Physics MCQ AIEEE 2003 2026 | ExamTest.live

mediumMCQAIEEE 20032026Physics

1 mark

The time period of a satellite of earth is 5 hours . If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become:

Solution & Step-by-step Explanation

According to Kepler's third law: T^{2} \propto r^{3} ⟹ T \propto r^{3/2} Given T_{1} = 5 hours and r_{2} = 4 r_{1} : \frac{T _{2}}{T _{1}} = (\frac{r _{2}}{r _{1}})^{3/2} T_{2} = 5 \times (4)^{3/2} = 5 \times (4)^{3} = 5 \times 2^{3} = 5 \times 8 = 40 hours

Try it yourself before checking the explanation above.

The time period of a satellite of earth is 5 hours . If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become:

10 hours

80 hours

40 hours

20 hours

Practice with timed mock tests and track your performance across Physics.