Q.A satellite of mass $1000\ \text{kg}$ orbits the Earth in a circular path of radius $6700\ \text{km}$ (i.e. at a height of $300\ \text{km}$). Taking $GM = 4.00 \times 10^{14}\ \text{m}^3/\text{s}^2$, find the total mechanical energy of the satellite in its orbit.

Given $GM = 4.00\times 10^{14}\ \text{m}^3/\text{s}^2$, $m = 1000\ \text{kg}$, $r = 6700\ \text{km} = 6.7\times 10^6\ \text{m}$.

$E = -\frac{GMm}{2r} = -\frac{(4.00\times 10^{14})(1000)}{2\times 6.7\times 10^6} = -\frac{4.00\times 10^{17}}{1.34\times 10^7} \approx -2.99\times 10^{10}\ \text{J}$ …

Substitute directly into $E = -GMm/2r$, keeping the orbital radius (not …

• Using the height ($300\ \text{km}$) rather than the full orbital radius ($6700\ \text{km}$, measured from the Earth's centre) in the formula. …