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Graviational pull

  • Thread starter aal0315
  • Start date
  • #1
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Homework Statement



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, excced the pull of the Earth? Ignore the effect of the sun in this calculation.

Homework Equations



F=G(M1M2/r^2)

The Attempt at a Solution



i dont know where to begin. does the force of the earth on the spaceship = the force of the moon on the spaceship and i solve for r?
 

Answers and Replies

  • #2
AEM
360
0

Homework Statement



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, excced the pull of the Earth? Ignore the effect of the sun in this calculation.

Homework Equations



F=G(M1M2/r^2)

The Attempt at a Solution



i dont know where to begin. does the force of the earth on the spaceship = the force of the moon on the spaceship and i solve for r?
You're on the right track. You want to find the distance from the earth at which the force on the spaceship by the earth equals the force on the spaceship by the moon. You will need to write the correct force law for each case. That is one force law with earth and spaeship masses and one with moon and spaceship masses. The you will have to figure out how to represent the distance so that you have only one variable to solve for when you set those two forces equal. I'd let R = the average distance between the earth and the moon (you can look it up) and r the distance from the earth to the space ship. Then what is the distance from the spaceship to the moon? Set up your force equations, set them equal, watch stuff cancel out and then solve for r. It will be a quadratic equation so be careful. You may need the quadratic formula. I don't know for sure because I haven't completed the algebra.
 
  • #3
37
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You need to break it into two parts.
Force on the rocket from the earth
Force on the rocket from the moon
When these two forces are equal, then the net force acting on the rocket will be zero

To solve this you can write each of the first two parts in terms of the equation you gave, using m1 as the mass of the rocket. For the first equation, m2 would be the mass of the earth, and r the distance from the earth to the rocket. For the second equation m2 would be the mass of the moon, and r the distance from the rocket to the moon

Set these two equations equal and solve for r to get the point where the space ship will feel no force. Once beyond this point (closer to the moon), the rocket will feel a stronger pull from the moon.
 
  • #4
41
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so i have
G(MshipMearth/r) = G(MshipMmoon/r)
Mearth/r = Mmoon/r (because G and Mship cancel each other out.
but if i try to solve for r, wont they just cancel out as well?
 
  • #5
cjl
Science Advisor
1,819
398
No, because the R is different in each case. The R for the moon case is the distance between the ship and the moon's center of gravity, while the R for the earth case is the R between the ship and the earth's center of gravity.
 
  • #6
41
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so would it be Mearth / (Rearth + r) = Mmoon / (Rmoon +r)
or Mearth / (distance b/w earth and moon - r) = Mmoon / (distance b/w earth and moon - r)?
i know that the distance between the earth and the moon is 384x10^3 km

i dont know what im doing at all. any help would be great.
 
  • #7
AEM
360
0
so would it be Mearth / (Rearth + r) = Mmoon / (Rmoon +r)
or Mearth / (distance b/w earth and moon - r) = Mmoon / (distance b/w earth and moon - r)?
i know that the distance between the earth and the moon is 384x10^3 km

i dont know what im doing at all. any help would be great.
Go back to my post. I told you how to write the r in each case. Again, (1) let the erath-spaceship distance be r. (2) let the moon-spaceship distance be (R-r) where capital R is the average earth-moon distance. Plug those into the expression for the force of gravity. That is the only way you will end up with an equation in 1 unknown --the r that you want.
 
  • #8
41
0
got the answer 1.56x10^7
Sounds right?
 
  • #9
41
0
think I got the calculations wrong. Now I got 3.79x10^5
 
  • #10
41
0
so the above answers are wrong because i know i never squared the r and R-r
i have tried the calculations over and over and get this huge number that is bigger than the distance of the earth to the moon. can someone help me?
Me(R^2) - (Me)(2)(R)(r) - (Me)(r^2) = Mm (r^2)
(5.98x10^24)(384x10^6)-2(5.98x10^24)(384x10^6)r-(5.98x10^24)r^2=7.35x10^22 r^2
8.82x10^41 - 4.59x10^33r + 5.98x10r^2 = 7.35x10r^2
using the quadratic equation
4.59x10^33 ± √[[(4.59x10^33)^2 - 4(5.91x10^24)(8.82x10^41)]/2(5.91x10^24)]
i get 4.59x10^33 which is too big. can anyone help please
 

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