Question

A satellite revolves around the earth at a height h1 = 2Re. Calculate the height h2 of the satellite from the surface of the earth at which the speed of the satellite is 3/2v1 (in a radical), where v1 is the speed of the satellite at the height h1.

Answer 1

From the conservation of energy:

$$0.5mv_1^2-\frac{GmM}{2R_e}=0.5m(1.5v_1)^2-\frac{GmM}{h_2}$$

$$GM\left(\frac{1}{h_2}-\frac{1}{2R_e}\right)=0.625v_1^2$$

$$\frac{1}{h_2}=\frac{1}{2R_e}+\frac{0.625v_1^2}{GM}$$

$$h_2=\left(\frac{1}{2R_e}+\frac{0.625v_1^2}{GM}\right)^{-1}$$