Question

4.
The period of revolution of an earth's
satellite close to the surface of earth is
60 minutes. The period of another earth's
satellite in an orbit at a distance of three
times earth's radius from its surface will
be (in minutes)
1) 90 2)90x √8 3) 270 4)480
​

Answer 1

Answer:

Explanation:

Distance of satellite from the center of earth is 4 times

T² α r³

T α r³/²

Time period of second satellite will become (4)3/2 ie 8 times

Hence T′=8T=8×60=480min

Hence 3 is the correct answer.