Energy to reach the moon

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.

• You forgot to multiply by 930? Sep 20, 2014 at 19:44
• Keep your units. The units for $\frac{GM}{r}$ would be $\frac{m^2}{s^2}$, not units of energy. That would be your way to see that the formula you are using is not giving you the answer you are expect. Sep 20, 2014 at 20:01
• @BowlOfRed - Concur wholeheartedly. Failing to do so is a very, very common beginner mistake. Sep 20, 2014 at 20:08
• @RobJeffries thats the distance from point P to Earth centre Sep 21, 2014 at 10:59
• Yes, that's in the diagram. I assume you get that from making the total gravitational field equal zero? You've edited your question so that several of the comments made on the original now make no sense. You are now within a gnats whisker of your required solution- what are you asking for? You can get closer by including the GPE due to the moon. Sep 21, 2014 at 11:48

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.

In this problem you need to consider the difference between the G.P.E. of the mass on the Earth's surface and the G.P.E. at this intermediate point. You haven't said how you have calculated this, but it seems likely that you have made the total gravitational field equal to zero at point P.

Your attempted solution is incomplete because you have forgotten the influence of the moon on the gravitational potential energy at point P.

The plot below shows the contributions to the GPE from the earth (blue), moon (green) and the sum of these (red).

As an aside, if you really want the solution to this problem to 3 sig figs then you need more more precise inputs for the Earth/Moon mass and separation (this varies of course) and to appreciate that because of centrifugal forces, the potential (i.e. the sum due to gravity and rotation) doesn't quite look like this and the point P is closer to the Earth at something called the L1 Lagrangian point: http://en.wikipedia.org/wiki/Lagrangian_point .

• There's no need for approximations here, nor for the L1 point. This is an introductory level question. At that level, the point between the Earth and the Moon at which the gravitational force between the two exactly cancel is deemed to be of special importance. That point is easy to calculate (it's a simple quadratic) and is easy for introductory students to understand. That this so-called zero point has zero meaning beyond introductory physics, well, introductory level physics classes don't discuss that. Sep 20, 2014 at 21:40
• Well, so is the potential energy of a 930 kg object on the surface of the Earth. That value of -58.7 GJ is clearly wrong to three decimal places of accuracy. Algebra-based physics classes (and this is almost certainly where this problem arises) take certain liberties with physics. How could you possibly teach the concept of the synodic frame to someone who struggles with the quadratic equation? Sep 20, 2014 at 22:16
• Ideally, yes. In practice, no. You have to realize the these kinds of problems are aimed at students who think that the quadratic equation represents the pinnacle of difficult mathematics. Sep 20, 2014 at 23:15
• We are not meant to use L1 points or rotational KE. I don't understand how you get your answer - where is the 0.11 and why does the energy required to bring an object in from infinity matter? Surely you only have to overcome the energy required to bring the object in from point P? And @DavidHammen this physics problem is not aimed at 14 year olds, but 17/18 in the English school system. Sep 21, 2014 at 9:42
• @Cobbles - Are you taking an algebra-based or calculus-based physics class? My previous comment was aimed more at advanced physics students who are predisposed to invoking general relativity when helping introductory level students solve a block sliding down a ramp problem. After all, "we shouldn't teach incorrect physics". Sep 21, 2014 at 10:09