The space shuttle is in a 300 km-high circular orbit

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mookie84
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Homework Statement



The space shuttle is in a 300 km-high circular orbit. It needs to reach a 610 km-high circular orbit to catch the Hubble Space Telescope for repairs. The shuttle's mass is 7.00×10^4 kg.

Homework Equations


½(-G(M-earth)(M-telescope))(1/r2 –1/r1)
G=6.67*10^6
Mass of earth=5.98*10^24
Mass of telescope= 7.0*10^4 kg
Radius of earth= 6.98*10^6 m


The Attempt at a Solution


What I did was plug in the numbers to the equation. For r2 I added the radius of Earth + the height needed to be attained (610km). For r1 I added the radius of Earth plus the original height (300km). I keep getting 8.4*10^10 which is the incorrect answer. What am I doing wrong?
 
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Sorry about that?
How much energy is required to boost it to the new orbit?
 
The difference in height is small compred to Earth's radius - what does "m g h" give?
 
which mass are you referring too? Earth or the satellite?
 
The shuttle - it's just the same question as lifting a weight on earth.
 
I've tried that it doesn't work. The question is asking me how many joules does it take to move the satellite from 310 km above the surface of Earth to 600 km. When i tried the equation you offered I'm neglecting the Earth's radius and mass which affects the gravitational pull
 
You are taking the radius and mass of the Earth into account in the factor 'g'
g = GM_earth/r_earth^2 = 6.67E-11 * 6E24 / (6.4E6 ^ 2) = 9.8 m/s^2

Potential energy is GMm/r so for the difference in height of 300km and 610km we have

dPE = GMm * ( 1/(r+300km) - 1/(r+610km) )
= 6.67E-11 * 6E24 * 7E4 * ( 1/(6370,000+300,000) - 1/(6370,000+610,000) )
= 6.67E-11 * 6E24 * 7E4 * 6.6E-9 = 1.8 E 11 J

or assuming change in gravity between 300km and 610 km is small and mass of shuttle is small compared to earth.
= m g h = 7E4 * 9.8 * 310,000 = 2.1 E11 J

Since the mass of the shuttle is only given to 1 sig figure the answer 2E11J is the same for both.



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