A Level Physics 9702: Total Energy of an Orbiting Satellite

A satellite of mass $m$ orbits a planet of mass $M$ in a circular orbit of radius $r$ .

(a) By equating the gravitational force on the satellite to the centripetal force required for its circular motion, show that the kinetic energy of the satellite in orbit is given by $E_k = \frac{GMm}{2r}$ [3]

(b) State the expression for the gravitational potential energy $E_p$ of the satellite, and hence show that the total energy of the satellite in its orbit is $E = -\frac{GMm}{2r}$ [2]

(c) A satellite of mass $600\text{ kg}$ orbits a planet of mass $5.00\times10^{24}\text{ kg}$ in a circular orbit of radius $7.00\times10^6\text{ m}$ . Take $G=6.67\times10^{-11}\text{ N m}^2\text{ kg}^{-2}$ . Calculate the total energy of the satellite in this orbit. [3]

(d) The satellite is moved to a new circular orbit of radius $1.40\times10^7\text{ m}$ , i.e. twice the radius in (c). By considering how the total energy depends on orbital radius, determine the new total energy without recalculating $GMm/2r$ from scratch, and state whether the total energy of the satellite has increased or decreased. [2]

Part (a): Deriving the kinetic energy

For the satellite to move in a circular orbit of radius $r$ at speed $v$ , the gravitational force must equal the centripetal force: $\frac{GMm}{r^2} = \frac{mv^2}{r}$

Cancelling $m$ and one factor of $r$ : $v^2 = \frac{GM}{r}$

Substituting into the definition of kinetic energy, $E_k=\tfrac{1}{2}mv^2$ : $E_k = \frac{1}{2}m\left(\frac{GM}{r}\right) = \frac{GMm}{2r}$

as required.

Part (b): Total energy of the orbit

The gravitational potential energy of the satellite is: $E_p = -\frac{GMm}{r}$

The total energy is the sum of kinetic and potential energy: $E = E_k + E_p = \frac{GMm}{2r} + \left(-\frac{GMm}{r}\right) = \frac{GMm}{2r} - \frac{2GMm}{2r} = -\frac{GMm}{2r}$

as required.

Part (c): Total energy for the given satellite

With $M=5.00\times10^{24}\text{ kg}$ , $m=600\text{ kg}$ , $r=7.00\times10^6\text{ m}$ : $GM = (6.67\times10^{-11})(5.00\times10^{24})$

$6.67\times5.00=33.35$ , and $10^{-11}\times10^{24}=10^{13}$ , so $GM=3.335\times10^{14}\text{ m}^3\text{ s}^{-2}$ .

$GMm = 3.335\times10^{14}\times600 = 3.335\times10^{14}\times6.00\times10^2 = 2.001\times10^{17}$

$2r = 2\times7.00\times10^6 = 1.40\times10^7\text{ m}$

$E = -\frac{GMm}{2r} = -\frac{2.001\times10^{17}}{1.40\times10^7} = -1.4293\times10^{10}\text{ J} \approx -1.43\times10^{10}\text{ J}$

Check by recomputing differently: $2.001/1.40=1.4293$ , and $10^{17}/10^7=10^{10}$ , giving $E=-1.4293\times10^{10}\text{ J}$ , matching exactly.

Part (d): Total energy at twice the radius

Since $E=-GMm/2r$ , the total energy is inversely proportional to $r$ (with $GM$ and $m$ unchanged, as it is the same satellite and planet). Doubling $r$ therefore halves the magnitude of $E$ : $E_{\text{new}} = \frac{E}{2} = \frac{-1.4293\times10^{10}}{2} = -7.146\times10^9\text{ J} \approx -7.15\times10^9\text{ J}$

Check by recalculating directly: $E_{\text{new}}=-\dfrac{GMm}{2(1.40\times10^7)}=-\dfrac{2.001\times10^{17}}{2.80\times10^7}=-7.146\times10^9\text{ J}$ , matching the halved value exactly.

Since $-7.15\times10^9\text{ J}$ is less negative than $-1.43\times10^{10}\text{ J}$ (it is closer to zero), the total energy of the satellite has increased. Energy had to be supplied to move it to the higher orbit, consistent with the total energy becoming less negative as orbital radius increases.

Final answers

(a) $E_k=\boxed{GMm/2r}$ (derived from equating gravitational and centripetal force, then substituting $v^2=GM/r$ into $\tfrac{1}{2}mv^2$ )
(b) $E_p=-GMm/r$ , and $E=E_k+E_p=\boxed{-GMm/2r}$
(c) $E = \boxed{-1.43\times10^{10}}\text{ J}$
(d) $E_{\text{new}} = \boxed{-7.15\times10^9}\text{ J}$ . The total energy has increased (become less negative)