Gravitation Field between the Earth and Moon

  • #1

Homework Statement


There is a point on the line between the centres of the earth and the moon where their gravitational fields have equal magnitude but are in opposite directions, effectively creating a point of zero gravity. Calculate the distance of this point from the centre of the earth.

gearth = 9.81 ms[itex]^{}-2[/itex]
mearth = 6.02 x10[itex]^{}24[/itex] kg

Homework Equations



F = GMm/d[itex]^{}2[/itex]

g = -GM/d[itex]^{}2[/itex]

The Attempt at a Solution


Using ratio of the inverse square laws Rm[itex]^{}2[/itex]/Re[itex]^{}2[/itex] = Mm/Me
Using F = 0 and gearth = gmoon
 

Answers and Replies

  • #2
tiny-tim
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hi rocketgirl93! :smile:

(try using the X2 and X2 buttons just above the Reply box :wink:)
gearth = 9.81 ms−2
mearth = 6.02 x1024 kg
you'll need a lot more data than that :confused:
 
  • #3
Thanks for the hint!

Ok, I'll try find some more, its from this horrific textbook where they dont give you all the constants/data you need at the start of the questions and its scattered throughout the chapter.
 
  • #4
HallsofIvy
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Thanks for the hint!

Ok, I'll try find some more, its from this horrific textbook where they dont give you all the constants/data you need at the start of the questions and its scattered throughout the chapter.
You mean they expect you to work? How awful!
 
  • #5
I like Serena
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Hey rocketgirl93! :smile:

Suppose you sky rocketed to this mysterious point...
Shall we give it a name?
Let's call it... L1!
(Yes it's actually called L1. :wink:)

Here's a neat picture!
330px-Lagrange_points2.svg.png



Oh, and just to get this started, how strongly would the moon pull at you?
 
  • #6
I take Physics, I'm not afraid to work! It makes it difficult because you dont know what you have to find first and how they want you to approach the questions.

I've found these;

Orbital radius for the Moon: 3.84 x108m
Mass of the Moon: 7.35 x1022kg
Actual Radius of the Earth: 6400km
 
  • #7
Hey rocketgirl93! :smile:

Suppose you sky rocketed to this mysterious point...
Shall we give it a name?
Let's call it... L1!
(Yes it's actually called L1. :wink:)

Here's a neat picture!
330px-Lagrange_points2.svg.png



Oh, and just to get this started, how strongly would the moon pull at you?
Hi! Would it pull at you at about 1.64 m/s2? Or at 1.64 x mass?
 
  • #8
I like Serena
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Good!

Looks like you have all the ingredients you need!
Your attempt was already in the right direction.

So what do you think you should do next?

Edit: how did you arrive at 1.64 m/s2?
 
  • #9
D H
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Suppose you sky rocketed to this mysterious point...
Shall we give it a name?
Let's call it... L1!
(Yes it's actually called L1. :wink:)
No, it's not.
 
  • #10
I like Serena
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No, it's not.
Oops. You're right.
I forgot for a moment about the centripetal force.

Ah well, it's still a nice picture!
 
  • #11
D H
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I take Physics, I'm not afraid to work! It makes it difficult because you dont know what you have to find first and how they want you to approach the questions.

I've found these;

Orbital radius for the Moon: 3.84 x108m
Mass of the Moon: 7.35 x1022kg
Actual Radius of the Earth: 6400km
You don't need the radius of the Earth to solve this problem. You do need the mass of the Earth. You also need one very important item which you have left out. Hint: Newton's universal law of <what>?
 
  • #12
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You don't need the radius of the Earth to solve this problem. You do need the mass of the Earth. You also need one very important item which you have left out. Hint: Newton's universal law of <what>?
Wouldn't the ratio of the inverse square laws do the trick?
 
  • #13
tiny-tim
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You don't need the radius of the Earth to solve this problem.
since she has g, she needs re if she doesn't have G :wink:
 
  • #14
D H
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Wouldn't the ratio of the inverse square laws do the trick?
What is this inverse square law to which you are referring? The original poster hasn't said anything about an inverse square law yet.
 
  • #15
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What is this inverse square law to which you are referring? The original poster hasn't said anything about an inverse square law yet.
It's in the attempt at a solution in the OP.
 
  • #16
D H
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Ah. I see that now.

So, another hint: If the point is at some distance d from the center of the Earth, how far is it from the center of the Moon?
 
  • #17
Good!

Looks like you have all the ingredients you need!
Your attempt was already in the right direction.

So what do you think you should do next?

Edit: how did you arrive at 1.64 m/s2?
I didnt know what g for the moon was but I remembered reading somewhere that it is about one sixth of the earths g (9.81) so i calculated this figure and guessed


You don't need the radius of the Earth to solve this problem. You do need the mass of the Earth. You also need one very important item which you have left out. Hint: Newton's universal law of <what>?
Of Gravitation, I posted that in my original solution

since she has g, she needs re if she doesn't have G :wink:
I do have G: 6.67x10^-11

Ah. I see that now.

So, another hint: If the point is at some distance d from the center of the Earth, how far is it from the center of the Moon?
radius of the moon + distance between earth and moon + distance to point from surface of the earth
 
  • #18
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When you are on surface of the earth, force exerted by the earth,
F=ma= GMearthm/r2earth

F=6.6742x10-11 x 5.97x1024m/(6.38x106)2
F=m(9.7888) m/s2

The moon also exerted the same force according Newton's Gravitational law.
You find the radius where the forces are equal.
 
  • #19
D H
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radius of the moon + distance between earth and moon + distance to point from surface of the earth
No! Draw a picture.

The distance between the Earth and Moon you cited in post #6 is center-to-center. You don't need the radius of the Earth or the Moon here.
 
  • #20
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I didnt know what g for the moon was but I remembered reading somewhere that it is about one sixth of the earths g (9.81) so i calculated this figure and guessed

Of Gravitation, I posted that in my original solution

I do have G: 6.67x10^-11
Note the difference between g and G.
g is the acceleration of gravity at the surface of the earth.
G is the universal constant of gravity.
For this calculation you will not need g and if you want you can use G (which is in your formula), but you can also use your ratio of the inverse square law in which case you won't need G.


radius of the moon + distance between earth and moon + distance to point from surface of the earth
You seem to be mixing the radius of the earth itself and the distance of the equilibrium point to the center of the earth.
Both can be written as re or Re, but they are very different things.

For your problem you won't need the radius of either earth or moon.
But you do need to find the distance of the equilibrium point to the center of the earth, and also the distance of the equilibrium point to the center of the moon.


Using ratio of the inverse square laws Rm2/Re2 = Mm/Me
This is what you already wrote down and what you can use.
Note that Re and Rm are NOT the radius of either earth or moon.
 
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