Space shuttle releases a sattelite into a circular orbit

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SUMMARY

The space shuttle must be moving at 7520 m/s relative to the Earth's center when it releases a satellite into a circular orbit 680 km above the Earth. The correct calculation involves using the formula for orbital velocity, v = sqrt(G * m(earth) / r), where r is the sum of Earth's radius (6380 km) and the altitude (680 km), resulting in r = 7060 km. A common mistake is using incorrect units, such as kilometers instead of meters, which can lead to significant errors in calculations.

PREREQUISITES
  • Understanding of gravitational force and orbital mechanics
  • Familiarity with the formula for orbital velocity
  • Knowledge of unit conversions, particularly between kilometers and meters
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Study the derivation of the orbital velocity formula, v = sqrt(G * m(earth) / r)
  • Learn about the implications of altitude on satellite velocity
  • Explore common pitfalls in physics calculations, particularly unit conversions
  • Review examples of satellite launches and their required velocities
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Aerospace engineers, physics students, and anyone interested in satellite dynamics and orbital mechanics will benefit from this discussion.

jmcmillian
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The space shuttle releases a sattelite into a circular orbit 680 km above the Earth. How fast must the shuttle be moving (relative to the Earth's center) when the release occurs?


I know that G (m*m(earth)/r^2) = m (v^2/r)...with r = r(earth) + h. In this equation, r=6380+680=7060 km. There also a manipulation of the given equations in the examples for velocity, sqr(G*m(earth)/r))

I am probably thinking along the wrong lines, but shouldn't the shuttle be moving at the same speed needed to keep the satellite in orbit?

So that is my approach to the problem thus far - trying to find the velocity of the satellite and equating it to the velocity of the shuttle upon release. To do that, I've used the sqr(G*m(earth)/r)) equation, which gives me:

sqr((6.67*10^-11)(5.98*10^24)/(7060))=237690 m/s. This isn't even close to the given answer in the book, 7520 m/s.
 
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jmcmillian said:
The space shuttle releases a sattelite into a circular orbit 680 km above the Earth. How fast must the shuttle be moving (relative to the Earth's center) when the release occurs?I know that G (m*m(earth)/r^2) = m (v^2/r)...with r = r(earth) + h. In this equation, r=6380+680=7060 km. There also a manipulation of the given equations in the examples for velocity, sqr(G*m(earth)/r))

I am probably thinking along the wrong lines, but shouldn't the shuttle be moving at the same speed needed to keep the satellite in orbit?

So that is my approach to the problem thus far - trying to find the velocity of the satellite and equating it to the velocity of the shuttle upon release. To do that, I've used the sqr(G*m(earth)/r)) equation, which gives me:

sqr((6.67*10^-11)(5.98*10^24)/(7060))=237690 m/s. This isn't even close to the given answer in the book, 7520 m/s.
What should the units be for R...I am getting an answer very close to what your book says.

Casey
 
Last edited:
Ah yes, it should be m instead of km. Often times I get caught up in the problem and lose focus in critical things such as units. Thanks for your help.
 

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