Question

The period of a satellite in a circular orbit of radius R is 12 hrs. The period of another satellite in a circular orbit of radius x is 24 hrs. Value of x is:-

Answer 1

Answer:

T=  

R

2π

​

[  

g

(R+h)  

3

 

​

]  

1/2

 

Here (R+h) changes from R to 4R. Hence period of revolution changes from T to (4  

3

)  

1/2

T=8T.

I think so

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Explanation:

Answer 2

Given : The period of a satellite in a circular orbit of radius R is 12 hrs.

The period of another satellite in a circular orbit of radius x is 24 hrs.

To Find :  Value of x is:-

Solution:

T=2π √R³/GM

The period of a satellite in a circular orbit of radius R is 12 hrs.

12 = 2π √R³/GM

 The period of another satellite in a circular orbit of radius x is 24 hrs.

24 = 2π √x³/GM

2 = √x³/R³

=> 4  = x³/R³

=> x³ = 4R³

=> x = ∛4 R

Value of x is ∛4 R

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