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Problem about orbital mechanics

Homework Statement
A satellite is in circular orbit around Earth. The orbit is such that the change in gravitational potential energy of the satellite-Earth system in going from the satellite’s location on the surface of Earth to its orbit height is equal to three times the satellite’s kinetic energy while in this orbit. How high above the surface of Earth (radius = R) is the satellite? :

A) 1⁄2R
B) 2⁄3R
C) R
D) 3⁄2R
E) 2R
Homework Equations
I think:
Ug= -GMm/r
Ke= 1/2 mv^2

V^2=(GM/r)

Umechanical= - 1/2 GMm/r
I tried it, but I am not getting no of the given answers

According to the statement, it is saying that

3 KE (in the orbit ) = ΔUg

So, beeing R the radius of the earth and R2 the radius of the orbit:

3 (1/2)(GMm/r2) = -GMm/r2 - (-GMm/R)

Canceling out the GMm:

(3/2)(1/r2)= (-1/r2) + (1/R)

Solving for R2

(3/2)(1/r2) + (1/r2)=(1/R)
(5/2)(1/r2)=(1/R)
(2/5)(r2)=R
r2= (5/2)R

Hoewever, this would be wrong, this is not a choice. Could someone teach me what I am doing wrong? thank you in advance :) .
 

PeroK

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How have you calculated the Kinetic Energy of the satellite?
 
Hello, with KE=3 (1/2)(mv^2)
where V^2 is GMm/r2
 

TSny

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So the answer would be
R2-R = (5/3)R- 1 R=(3/2) R
 

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