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  • #1
The mass of the moon is 0.0123 times that of the Earth. A spaceship is traveling along a line connecting the centers of the Earth and Moon. At what distance from the Earth does the spaceship find the gravitational pull of the Earth equal in magnitude to that of the Moon? Express your answer as a percentage of the distance between the centers of the two bodies.

I really don't understand how to do this someone help please.
 

Answers and Replies

  • #2
LeonhardEuler
Gold Member
859
1
From Newtons law of gravity, the force, F, from one body is given by:
[tex]F=\frac{Gm_1m_2}{r^2}[/tex]
G being the gravitational constant, m1 and m2 being the masses of the body making the force and the body recieving the force, and r being the distance between. Make m1 the mass of the ship. Now you require the force from the moon and the force from the earth to be equal. Do you see what equation to set up?
 
  • #3
Ok Where do I get the mass of the ship from all the information I am given I have already wrote above. And I don't understand how to make the forces be equal. Please help.
 
  • #4
574
0
You don't need the mass of the spaceship to solve the problem. When you equalize the forces excerted by the moon and earth the mass of the spaceship can be divided off the equation leaving the distances as the unknowns.
 
  • #5
LeonhardEuler
Gold Member
859
1
Keep the mass of the ship as m1. It will eventually cancel its way out of the question. So will G. To find out when they are equal, find equations for the forces from each of the bodies and set them equal to each other. If d is the distance between the moon & earth and x is the distance btween the earth and ship, then (d-x) gives the distance between the moon & ship. Gotta go, good luck!
 
  • #6
286
0
inha said:
You don't need the mass of the spaceship to solve the problem. When you equalize the forces excerted by the moon and earth the mass of the spaceship can be divided off the equation leaving the distances as the unknowns.
Or in other words, you can make the mass of the spaceship whatever you like. Personally I would choose mass = 1.

Why you can do this? Because the mass of the ship doesn't change. You can find out the actual mass of the ship, or make it mass 1,000 or 1,000,000. No matter though; the point between the earth and the moon where the moon's gravity equals the earth's gravity will be the same no matter what the ship's mass is because it doesn't change.
 
  • #7
Ok once again I'm lost you end up with r^2 canceling out too I cannot find anyother way to set it up
 
  • #8
574
0
total distance between earth and moon: d. distance from earth to the spaceship: x. distance between moon and the spaceship d-x. Can you figure it out now?
 
  • #9
286
0
lilkrazyrae said:
Ok once again I'm lost you end up with r^2 canceling out too I cannot find anyother way to set it up
inha's hint is very useful. You assumed that the two measures of radius are equal when in fact they are not. Well...they are at d/2 (according to inha's variables).

Initially we told you that

[tex]F_{gravity}=\frac{Gm_1m_2}{r^2}[/tex]

So it follows that:

[tex]F_{earth}=\frac{Gm_{ship}m_{earth}}{r^2}[/tex]

where r equals the distance from the earth to the ship. And it also follows that:

[tex]F_{moon}=\frac{Gm_{ship}m_{moon}}{r^2}[/tex]

but the r in this equation is the distance from the moon to the ship.

You want to find out when these two equations equal each other, or in other words:

[tex]F_{moon}= F_{earth}[/tex]

So begin by setting their equations equal to each other (I'm sure you've already done this, but pay attention to the variables...same letters but different subscripts):

[tex]\frac{Gm_{ship}m_{moon}}{r_{m}^2}=\frac{Gm_{ship}m_{earth}}{r_{e}^2}[/tex]

where [itex]r_{m}[/itex] is the distance between the ship and the moon, and [itex]r_{e}[/itex] is the distance between the ship and the earth. Remember that the mass of the moon is only 0.0123 the mass of the earth. That expresses the mass of the moon in terms of the mass of the earth, so we can eliminate the [itex]m_{moon}[/itex] and express it in earth masses. That means:

[tex]\frac{Gm_{ship}0.0123m_{earth}}{r_{m}^2}=\frac{Gm_{ship}m_{earth}}{r_{e}^2}[/tex]

Notice how the variables start to drop off. G, the mass of the ship, and the mass of the earth m cancel out now:

[tex]\frac{0.0123}{r_{m}^2}=\frac{1}{r_{e}^2}[/tex]

Getting it? I hope I haven't done too much, lol! But I'm sure your confusion was because you weren't keeping in mind that, even though the variable letters all represent similar IDEAS, you should not assume that they represent the same QUANTITIES OF THOSE IDEAS in this problem. If you have two different distances you should use two different variables. In this case we use r, but qualify it further with a subscript. Heck, you don't have to use subscripts if you use an entirely different letter, but the r's plug in to the gravity equation nicely. It's fair enough to rewrite this equation a new way:

[tex]\frac{Gm_{ship}m_{moon}}{r_{m}^2}=\frac{Gm_{ship}m _{earth}}{r_{e}^2}[/tex]

[tex]\frac{Gab}{d^2}=\frac{Gae}{f^2}[/tex]

Where
a = mass of the ship
b = mass of the moon
d = distance between ship and moon
e = mass of the earth
f = distance between ship and earth

Notice that you don't run into the problem you had earlier, with the r^2 cancelling out here, because you can't cancel f's with d's!!
 
Last edited:
  • #10
Sorry dumb mistake!! Ok just one more thing I think I did everything right and I got 9.02d=f the "Express your answer as a percentage of the distance between the centers of the two bodies" is throwing me off.
 

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