# Gravitation - exam in 2 days

1. May 12, 2010

### mlostrac

1. The problem statement, all variables and given/known data

A spaceship is launched and starts moving directly towards the Moon. At what distance from the Earth will the pull of the Moon, on the spaceship, exceed the pull of the Earth? Ignore the effect of the sun in this calculation.

2. Relevant equations

F=G(M1M2/r^2)

3. The attempt at a solution
I have no idea where to start. I need a push in the right direction :s

2. May 12, 2010

### bjd40@hotmail.com

the eqn is right. u have to know the distance of moon from earth. it must be given or u have to look up the table for it. assuming it to be x and the distance of the spaceship from earth to be r., calculate the pull on the spaceship by earth and by the moon seperately (the distance this time wopuld be x - r). u have to know the mass of earth and moon also. equate them and get the result for r. at that value of r both the attractions are same. now it will be easy.

3. May 12, 2010

### mlostrac

Thanks for the help. Here's what I got; look right?

Constants:
Me (Mass of Earth) = 5.98 x 10^24 kg
Mm (Mass of Moon) = 7.35 x 10^22 kg
z (Distance from Earth to Moon) = 3.84 x 10^8 m

G(Me x Mss)/ R^2 = G(Mm x Mss)/ (z-R)^2

z^2 (Me) = R^2 + R^2 (Mm)

x^2 = 1.01R^2

R = 3.82 x 10^8

So this means that once the ship gets past 3.82 x 10^8 m from the earth, the pull of the moon is greater?

4. May 13, 2010

### Jokerhelper

you should always do a check to see if your answer is reasonable. According to your answer, once the distance from the Earth is greater than 3.82 x 108 m, the gravitational pull of the Moon will be greater than the Earth's pull. In this case, what would be a good way to verify your answer?

Last edited: May 13, 2010
5. May 13, 2010

### mlostrac

Hmmm. I know the force of gravity on the moon is 17% of that on earth (9.8 m/s^2 versus 1.66 m/s^2). I'm not exactly sure how to check my answer though...

Do I use Mss x g = G (Mss x Me)/ R^2 --> and calculate to find R for each (the spaceship from the Moon and the spaceship from the earth)?

:s

6. May 13, 2010

### Jokerhelper

no no no. let's get somethings straight. The gravity of a planet is always related to the distance of the object to such planet.
For example, have you ever wondered why you often use in high school 2 formulae to calculate the gravitational force of the Earth?

$$F_{g} = mg$$ and also $$F_{g} = G\frac{mM_{e}}{r^{2}}$$

Therefore, $$g = \frac{GM_{e}}{r^{2}}$$, for the earth's gravity. This is also applicable for any planet or object with mass. Note that the values of 9.81 m/s2 for the Earth and 1.66 m/s2 for the moon are true only at the surfaces of the respective bodies, where the radius is fixed. Hence, you cannot using those values to verify if your answer is correct.

By the way, according to your answer $$G\frac{M_{e}M_{ss}}{R^{2}} = G\frac{M_{m}M_{ss}}{(z-R)^{2}}$$ when R = 3.82x102 . Maybe you could plug in your known values to verify if this is true.

7. May 14, 2010

### mlostrac

I got 3.82 x 10^8 m (the distance from the earth to the spaceship) using my numbers. So am I wrong? I'm assuming the Earth will have a greater pull than the moon, so this answer seems like it makes sense in that respect. But I'm not sure if I did everything correctly? My exam's tonight (Friday), and I'd like to figure this out before then.

8. May 14, 2010

### Jokerhelper

woops, i meant to copy your answer 3.82 x10^8 m (not 10^2), which i think might be incorrect. your approach is correct, but I think you might have made a mistake while expanding (z-R)^2.

Remember (a+b)^2 = a^2 + 2ab + b^2.

Actually, I'll post what you should have done, since your exam is tomorrow and, as a student myself, i can relate to how it feels to not be able to solve a question right before an exam. Just give me a few minutes to work it out.

9. May 14, 2010

### mlostrac

wouldn't (z-R)^2 be (z-R) (z+R) = z^2 -zR+zR -R^2 therefore equaling z^2 - R^2 ? That's what I did originally

10. May 14, 2010

### Jokerhelper

no. (z-R)^2 =/= (z-R) (z+R)
but z^2 - R^2 = (z-R) (z+R)

remember, (z-R)^2 is just a modification of (a+b)^2, so use the FOIL trick.
you are confusing the square of a binomial with difference of squares. To see this, just pick simple numbers for z and R (ex 5 and 6) to verify this.

for some practice, http://www.sac.edu/homepages/leeds_kelvin/factoring/binomialsquare.html [Broken]

Last edited by a moderator: May 4, 2017
11. May 14, 2010

### Jokerhelper

Me(Mass of Earth) = 5.98 x 10^24 kg
Mm (Mass of moon) = 7.35 x 10^22 kg
z (Distance from Earth to Moon) = 3.84 x 10^8 m
Mss is the mass of the shuttle (although it is irrelevant)

G is the gravitational constant

We assume that at a certain distance from the Earth R the gravitational force exerted by the Earth will be equal to the moon's, and past this point, the moon's pull will be greater.

Thus:

$$F_{g Earth} = F_{g Moon}$$

$$G\frac{M_{e}M_{ss}}{R^{2}} = G\frac{M_{m}M_{ss}}{(z-R)^{2}}$$ as you did, which simplifies to
$$\frac{M_{e}}{R^{2}} = \frac{M_{m}}{(z-R)^{2}}$$

Therefore:
$$0 = M_{m}R^{2} - M_{e}(z-R)^{2}$$ and by expanding the square:
$$0 = M_{m}R^{2} -M_{e}(z^{2} - 2zR + R^{2})$$
$$0 = (M_{m} -M_{e})R^{2} + 2zM_{e}R - z^{2}M_{e}$$, so to solve for R you need to use the quadratic formula. try to take it from here

Last edited: May 14, 2010