# Homework Help: The space shuttle is in a 300 km-high circular orbit

1. Nov 23, 2007

### mookie84

1. The problem statement, all variables and given/known data

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.

2. Relevant 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

3. 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?

2. Nov 23, 2007

### mgb_phys

what is the question?

3. Nov 23, 2007

### mookie84

How much energy is required to boost it to the new orbit?

4. Nov 23, 2007

### mgb_phys

The difference in height is small compred to earth's radius - what does "m g h" give?

5. Nov 23, 2007

### mookie84

which mass are you referring too? earth or the satellite?

6. Nov 23, 2007

### mgb_phys

The shuttle - it's just the same question as lifting a weight on earth.

7. Nov 23, 2007

### mookie84

I've tried that it doesnt 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 earths radius and mass which affects the gravitational pull

8. Nov 23, 2007

### mgb_phys

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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