How fast must a satellite leave Earth's surface

  • Thread starter fizzyy
  • Start date
  • #1
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Homework Statement



How fast must a satellite leave Earth's surface to reach an orbit with an altitude of 895 km?

Homework Equations



v = √GM/r

The Attempt at a Solution



G = 6.67 x 10^-11
M = 5.98 x 10^24
r = (6.38 x 10^6) + (8.95 x 10^5) = 7.275 x 10^6

v = √(6.67 x 10^-11)(5.98 x 10^24)/(7.275 x 10^6)
v = 7404.5 m/s

But this is the speed the satellite needs to stay in orbit. Is this the same speed that it needs to leave the Earth's surface? Do I need to use v = √2GM/r instead to find the velocity it needs to escape the gravitational field at Earth's surface and go into orbit?
 

Answers and Replies

  • #2
HallsofIvy
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To the velocity required to orbit at that height, you need to add the velocity necessary so that the kinetic energy can be 'traded off' for the change in potential energy to get to that height.
 
  • #3
7
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Would that be

Ek = ΔEg
1/2mv^2 = -GMm/r2 - (-GMm/r1)

solve for v, then add it to 7404.5 m/s?
 
  • #4
7
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Ohhh wait, conservation of energy applies, right?

Ek1 + Eg1 = Ek2 + Eg2

just plug v2 = 7404.5 m/s in
 

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