Physics, asked by sujitpra10, 10 months ago

two satellites are launched in to cicular orbit of , with radius R and 1.02 around earth . Then percentage difference in time period of satellite is
(1)1%
(2)2%
(3)3/2%
(4)3%​

Answers

Answered by Anonymous
1

Answer:

According to Kepler's law,

 

⟹T  

2

∝R  

3

 

⟹T∝R  

(3/2)

     ......................equation (1)

where,

T= time period

R= radius

Now, differentiating both sides we get  

dT=(  

2

3

​  

)R  

(  

2

1

​  

)

dR

Now, Dividing both sides by equation (1), we get

⟹  

T

dT

​  

=  

R  

(  

2

3

​  

)

 

(  

2

3

​  

)R  

(  

2

1

​  

)

 

​  

dR

⟹  

T

dT

​  

=  

2

3

​  

 

R

dR

​  

      .............................equation (2)

Now,  

Percentage difference in radius, =  

R

dR

​  

×100%=  

R

1.02R−R

​  

×100%=2%

So, Percentage difference in time period , =  

T

dT

​  

×100=  

2

3

​  

×2=3%

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