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

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

Ek + Eg (earth) = 1/2 Eg

(could you please show how to substitute to solve for v)

this question is on the forum and it was suggested to add this v once calculated to 7404.5 m/s which was calculated from v= square root of GM/r.

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 Recognitions: Homework Help Welcome to PF. PE + KE = PE + KE What they want to know is the KE at launch to end in the PE and KE in orbit.
 could you please show how to substitute for this equation? for 1/2mv squared i do not know what to use for mass.

Recognitions:
Homework Help

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

Your PE at any point is given by -GMm/r. We will shorten that to -μ*m/r

μ = GMe is the standard gravitational parameter for Earth = 398,000 km³/s²
http://en.wikipedia.org/wiki/Standar...onal_parameter

This means that

-μm/Rearth + 1/2*m*V2launch = -μm/Rorbit + 1/2*m*Vorbit2

The mass of the object drops out.

V2launch = 2*(μ/Rearth -μ/Rorbit) + Vorbit2

 Thank-you so much!!

 Tags orbit, physics, satellite