A spacecraft is positioned on the line connecting the centers of Earth and the spacecraft

PHYSICS · Multi-part · CBSE · Physics · Gravitation · Gravitation

Problem

1.20\times10^{3}\,\text{kg}

spacecraft is positioned on the line connecting the centers of Earth and the spacecraft. Relative to the conventional zero of gravitational potential energy at infinite separation, the potential energy of the Earth–spacecraft system is measured to be

-3.0\times10^{10}\,\text{J}

at the spacecraft’s current location. The mass and mean radius of Earth are

M_E = 5.97\times10^{24}\,\text{kg}

and

R_E = 6.37\times10^{6}\,\text{m}

, respectively. Use

G = 6.67\times10^{-11}\,\text{N}\,\text{m}^2\,\text{kg}^{-2}

. (a) Determine the distance

r

between the center of Earth and the center of the spacecraft that corresponds to the given potential energy. (b) Calculate the spacecraft’s altitude above Earth’s surface at this location. (c) Briefly describe how choosing a different reference level for potential energy would affect (i) the numerical value of the potential energy and (ii) the numerical value of the distance found in part (a).

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