Mechanics> grav16...

For a satellite orbiting close to the surface of earth the period of revolution is 84 mins.The time period of another satellite orbiting at a hegiht three times the radius of earth from its surface will be

a) (84)2root[2] mins

b) (84) 8 mins

c) (84) 3root[3] mins

d) (84) 3 mins

Arnab Mandal , 14 Years ago

Grade 11

2 Answers

Ashwin Muralidharan IIT Madras

Hi Arnab,

The time period of revolution,

T α 1/√(R³). So T = k/√(R³)

Hence T1/T2 = √(R2³)/√(R1³)

Or T1/T2 = √(R³)/√(64R³) ------------(the new distance is 4R from the centre).

So T1 = T2/8.....

Hence T1 = 84/8 mins.

In Option (B), you should have 84/8 mins.

Hope that helps.

Best Regards,

Ashwin (IIT Madras).

Last Activity: 14 Years ago

Ashwin Muralidharan IIT Madras

Hi Arnab,

Kindly note that,

Time period is T is directly proportional to √(R³).

And not inversely propoertional.

Extremely sorry for the error before.

So new Timer period = 84*8 mins.

Option (B).

Hope this helps.

Best Regards,

Ashwin (IIT Madras).

Last Activity: 14 Years ago

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Mechanics> grav16...

For a satellite orbiting close to the surface of earth the period of revolution is 84 mins.The time period of another satellite orbiting at a hegiht three times the radius of earth from its surface will be a) (84)2root[2] mins b) (84) 8 mins c) (84) 3root[3] mins d) (84) 3 mins

Arnab Mandal , 14 Years ago

Ashwin Muralidharan IIT Madras

Ashwin Muralidharan IIT Madras

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