# Space shuttle energy problem

## Homework Statement

The space shuttle is in a 250 km high circular orbit. It needs to reach a 610 km high circular orbit to catch the Hubble telescope for repairs. The shuttle's mass is 75000 kg. How much energy is required to boost it to the new orbit?

## The Attempt at a Solution

I keep coming up with answer that is twice the answer in the back of my book

Uf= E + Ui
(-6.67E-11)(5.98E24kg)(75000kg)/(610000m+6.37E6) = E + (-6.67E-11)(5.98E24)(75000kg)/(250000m+6.37E6)

E= 2.33E11 J

book gives 1.17E11

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D H
Staff Emeritus
You are ignoring kinetic energy.

## Homework Statement

The space shuttle is in a 250 km high circular orbit. It needs to reach a 610 km high circular orbit to catch the Hubble telescope for repairs. The shuttle's mass is 75000 kg. How much energy is required to boost it to the new orbit?

## The Attempt at a Solution

I keep coming up with answer that is twice the answer in the back of my book

Uf= E + Ui
(-6.67E-11)(5.98E24kg)(75000kg)/(610000m+6.37E6) = E + (-6.67E-11)(5.98E24)(75000kg)/(250000m+6.37E6)

E= 2.33E11 J

book gives 1.17E11
In Uf= E + Ui,
Ui should be the sum of kinetic energy mv^2/2 and potential energy -GmM/r
using GmM/r^2 = mv^2/r, we can get mv^2/2 = GmM/(2r)
so U = -GmM/(2r), not -GmM/r as you used.

I know for a fact that this solution is correct:

We are looking for the change in mechanical energy.

Emech = (1/2)Ug
Therefore, ∆Emech = (1/2)∆Ug

G = 6.67 X 10^-11
Me = 5.98 X 10^24 kg
Ms = 7.5 X 10^4 kg
Re = 6.37 X 10^6 m
r1 = Re + 2.5 X 10^5 m
r2 = Re + 6.1 X 10^5 m

Solve for Ug at r1. (Call this Ug1)
Ug1 = (-G(Me)(Ms))/(r1) = -4.519 X 10^12 J

Solve for Ug at r2. (Call this Ug2)
Ug2 = (-G(Me)(Ms))/(r2) = -4.286 X 10^12 J

∆Ug = Ug2 - Ug1 = 2.33 X 10^11 J

∆Emech = (1/2)∆Ug = 1.17 X 10^11 J

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