Achieving an 895 km Orbit: Calculating v

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To achieve an 895 km orbit, a satellite must reach a specific velocity calculated using gravitational potential energy (PE) and kinetic energy (KE) equations. The standard gravitational parameter for Earth, μ, is 398,000 km³/s², which is essential for these calculations. The equation simplifies to V²launch = 2*(μ/Rearth - μ/Rorbit) + Vorbit², allowing for the determination of the launch velocity. The mass of the satellite cancels out in the equations, making it unnecessary for the calculations. Understanding these principles is crucial for successfully launching a satellite into the desired orbit.
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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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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.
 
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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/Standard_gravitational_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!
 

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