Question

A 50.0-kg satellite is in a circular orbit about a planet. the potential energy of the orbit is -4.00 × 107 j. what is the speed of the satellite?

Answer 1

1 answer · Physics 

 Best Answer

The relation between period and orbit radius of a satellite is 
T^2 = (4π^2/GMe) r^3 

4π^2/GMe 
= 4π^2 / [6.67x10^-11 x6.0x10^24)] 
= 0.9865 x 10^-13 
=> r^3 = (1/0.9865x10^-13) T^2 = (1.014 x 10^13) T^2 

=> r1 = [(1.014 x 10^13)(120x60)^2]^(1/3) = 8.07 x 10^6 m 
and r2 = [(1.014 x 10^13)(180x60)^2]^(1/3) = 10.57 x 10^6 m 

=> Mechanical energy needed 
= - GMem/2 [1/r2 - 1/r1] 
= 6.67 x 10^-11 x 6.0 x 10^24 * 50 * 10^-6 * [1/8.07 - 1/10.57] 
= 1000.5 x 10^7 * (0.1239 - 0.0945) 
= 2.94 x 10^8 J. 

[Answer ( a ) 2.9 x 10^8 J.] 

Edit: 
My student Neel Pradip Shah suggested the following alternative simpler method: 
v = 2πr/T ... ( 1 ) and mv^2/r = GMem/r^2 ... ( 2 ) 
Eliminating r from ( 1 ) and ( 2 ), 
v = (2πGMe/T)^(1/3) 

Mechanical Energy 
= - (1/2) mv^2 
=> minimum energy needed 
= - (1/2) mv2^2 - [(-(1/2)mv1^2] 
= (1/2) m [v1^2 - v2^2] 
= (1/2) m * (2πGMe)^(2/3) * [1/T1^(2/3) - 1/T2^(2/3)] 
= (1/2) * 50 * (2π * 6.67 x 10^-11 x 6.0 x 10^24)^(2/3) * [1/(120*60)^(2/3) - 1/(180*60)^(2/3)] 
= 25 * 10^8 * (184.91) * (0.002682 - 0.002047) 
= 2.93 x 10^8 J. 

[Answer ( a ) 2.9 x 10^8 J.]