Answer :
The spaceship reaches a height [tex]8.3781\times10^6m[/tex] from the center of the earth.
Conservation of Energy:
Since the gravitational force is a conservative force, the concept of conservation of energy will help in calculating the distance.
At the highest point, the velocity of the spaceship will be zero so the kinetic energy plus potential energy at the ground of the spaceship will be converted totally into gravitational potential energy.
Let the highest point be at a distance h from the surface of the earth and the mass of spaceship be m, let R be the radius of the earth, then:
[tex]\frac{1}{2}mv^2-\frac{Gm_{earth}m}{R}=-\frac{Gm_{earth}m}{R+h} \\\\\frac{1}{2}mv^2=\frac{Gm_{earth}m}{R}-\frac{Gm_{earth}m}{R+h} \\\\\frac{1}{2}mv^2=Gm_{earth}m}[\frac{1}{R} -\frac{1}{R+h} ]\\\\\frac{1}{2}mv^2=Gm_{earth}m}[\frac{h}{R(R+h)} ] \\\\\frac{R(R+h)}{h}=\frac{2Gm_{earth}}{v^2}\\\\\frac{R^2}{h}+R=\frac{2\times6.67428\times10^{-11}\times5.9742\times10^{24}}{5534^2}=26\times10^6m\\\\\frac{R^2}{h}=(26-6.3781)\times10^6m\\\\h=\frac{(6.3781)^2\times10^{12}}{19.62\times10^6}\\\\h=2.073\times10^6m[/tex]
So the distance from the center will be:
d = R+h
[tex]d = 8.3781\times10^6m[/tex]
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