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

In summary, the space shuttle needs to reach a 610 km-high circular orbit in order to catch the Hubble Space Telescope for repairs. The difference in height is small compared to Earth's radius, so it needs to use energy to reach the new orbit.
  • #1
mookie84
8
0

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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  • #2
what is the question?
 
  • #3
Sorry about that?
How much energy is required to boost it to the new orbit?
 
  • #4
The difference in height is small compred to Earth's radius - what does "m g h" give?
 
  • #5
which mass are you referring too? Earth or the satellite?
 
  • #6
The shuttle - it's just the same question as lifting a weight on earth.
 
  • #7
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
 
  • #8
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.



=
 

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

1. What is a circular orbit?

A circular orbit is when an object in space is moving around another object (such as a planet or star) in a circular path, with the center of the object being orbited at the center of the circle.

2. How high is a 300 km-high circular orbit?

A 300 km-high circular orbit means that the space shuttle is 300 kilometers (or about 186 miles) above the surface of the object it is orbiting.

3. How fast is the space shuttle moving in a 300 km-high circular orbit?

The speed of an object in a circular orbit depends on its altitude, but for a 300 km-high orbit, the space shuttle would be moving at a speed of approximately 7.8 km/s (or 17,500 mph).

4. How long does it take for the space shuttle to complete one orbit in a 300 km-high circular orbit?

The time it takes for an object to complete one orbit depends on its speed and the size of the orbit, but for a 300 km-high orbit, the space shuttle would take approximately 90 minutes to complete one orbit.

5. How is the space shuttle able to stay in a 300 km-high circular orbit?

The space shuttle is able to stay in a circular orbit due to the balance between its speed and the gravitational pull of the object it is orbiting. The shuttle's speed keeps it from falling towards the object, while gravity keeps it from flying off into space.

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