Space shuttle energy problem

  • Thread starter bcjochim07
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  • #1
bcjochim07
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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?


Homework Equations





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
 

Answers and Replies

  • #2
D H
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You are ignoring kinetic energy.
 
  • #3
SimonZ
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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?


Homework Equations





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.
then we can get the half of your answer.
 
  • #4
barbarsk
1
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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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