Achieving an 895 km Orbit: Calculating v

[phy_]

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.

 

[LowlyPion]

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.

 

[phy_]

could you please show how to substitute for this equation? for 1/2mv squared i do not know what to use for mass.

 

[LowlyPion]

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

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

This means that

-μm/R_earth + 1/2*m*V²_launch = -μm/R_orbit + 1/2*m*V_orbit²

The mass of the object drops out.

V²_launch = 2*(μ/R_earth -μ/R_orbit) + V_orbit²

 

[phy_]

Thank-you so much!