A spacecraft is in a circular parking orbit above Earth’s surface

PHYSICS · Single correct · CBSE · Physics · Gravitation · Gravitation

Problem

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

spacecraft is in a circular parking orbit

2.0\times10^{6}\,\text{m}

above Earth’s surface. Neglecting atmospheric drag, what is the minimum kinetic energy the spacecraft must be given (at that altitude) so that it can escape Earth’s gravitational influence? Constants:

M_{\text{E}}=6.0\times10^{24}\,\text{kg}

R_{\text{E}}=6.4\times10^{6}\,\text{m}

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

(a)

4.8\times10^{10}\,\text{J}

(b)

2.0\times10^{10}\,\text{J}

(c)

2.4\times10^{10}\,\text{J}

(d)

9.5\times10^{10}\,\text{J}

Related theory

Gravitational Potential Energy
Gravitation
Escape Velocity
Escape speed

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