Question

A satellite is put into orbit around the earth first by being carried on a rocket that climbes vertically to h = 217 km from the earth\'s surface and then being propelled at a velocity of v = 9.65 km/s. This initial velocity is perpendicular to the position vector from the earth\'s center to the satellite. What is the maximum distance from the earth\'s center the satellite will reach with a closed orbit around the earth?

Answer 1

i think this helps

Answer 2

According to the law of conservation of energy

$\frac{1}{2}mv^{2}+mg(R+h)=\frac{1}{2}mv_{0}^{2}+mgd$

but, $v_{o}^{2}=\frac{GM}{d}$

Then the above expression can be written as,

$\frac{1}{2}mv^{2}+mg(R+h)=\frac{1}{2}(\frac{GM}{d})+mgd$

$2gd^{2}-d[v^{2}+2g(R+h)]+GM=0$

By substituting the values in the above equation, we have

            d = 5.89 * 10^6 m