I've been stuck on this question for a couple of days now.
Given that the potential energy of a 930 kg object on the Earth's surface is -58.7GJ, calculate the minimum energy required for the 930 kg object to reach the Moon.
Mass of Earth: $6.0 \times 10^{24}kg$ Mass of Moon: $7.4 \times 10^{22} kg$
Diagram:
|EARTH|———$3.6\times10^8m$——–|P|—-$0.4\times10^8m$—-|MOON|
The minimum energy is the energy to get to point P because the Moons gravity will pull the rocket the rest of the way.
I don't understand why my answer is wrong:
$$Work\space done\space bringing\space 930kg\space from\space P\space to\space \infty = -\frac{GMm}{r}$$
$$ = -\frac{G\times6\cdot10^{24}\times930}{3.6\cdot10^8} = -1.03GJ$$
my answer = work done bringing 930kg from Earth to infinity – work done bringing 930kg from P to infinity = $58.7-1.03 = 57.67GJ$
But this is incorrect.
The answer is 57.5 GJ by the way.
Best Answer
You calculated the specific potential energy at that distance. You were asked to calculate the potential energy needed to reach that point.
You did two things wrong in that calculation. You forgot to multiply by 930 kg and you forgot to use the given condition "that the potential energy of a 930 kg object on the Earth's surface is -58.7GJ".
What you need to do is calculate the change in potential energy from that at the surface of the Earth to that at that special point.
As an aside, you should get in the habit of always carrying the units along with your calculations. You would have seen the error of forgetting to multiply by mass if you did your calculation as
$$V = -\frac {GM}r = - \frac {(6.674\times10^{-11}\,\text{m}^3\text{kg}^{-1}\text{s}^2) \,(5.972\times10^{24}\,\text{kg})}{3.6\times10^8\,\text{m}} = -1.1\times10^6 \text{m}^2/\text{s}^2$$
That doesn't have units of energy. It has units of velocity squared, or energy per unit mass.