Where between the Moon and the Earth is the gravitational potential=0 ?

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SUMMARY

The discussion focuses on identifying the point between the Earth and the Moon where the gravitational potential from both bodies is equal, utilizing the masses of Earth (5.98x1024 kg) and Moon (7.35x1022 kg), and the distance between them (3.84x108 m). Participants clarify that this point does not equate to zero gravitational field, as gravitational acceleration (g) is inversely proportional to the square of the distance (g=Gm/r2). The concept of apparent weightlessness in orbit is also discussed, emphasizing that true weightlessness occurs only when gravitational acceleration (g) is zero.

PREREQUISITES
  • Understanding of gravitational potential and field equations (V=GM/r, g=Gm/r2)
  • Knowledge of mass values for celestial bodies (Earth and Moon)
  • Familiarity with concepts of weightlessness and free fall
  • Basic understanding of Lagrange points and orbital mechanics
NEXT STEPS
  • Calculate the exact distance from Earth to the gravitational equilibrium point using gravitational potential equations.
  • Explore the concept of Lagrange points and their significance in celestial mechanics.
  • Research the conditions for true weightlessness in different gravitational fields.
  • Study the effects of gravitational acceleration on objects in orbit and how it relates to perceived weight.
USEFUL FOR

Students studying physics, particularly those interested in gravitational forces, orbital mechanics, and the principles of weightlessness in space travel.

shk
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Homework Statement



Somewhere between the Earth and the Moon there is a point where the gravitational potential due to the Earth exactly equals that due to the Moon.
i)At what distance from the Earth is this point?
Mass of Earth = 5.98x10^24 kg
Mass of Moon= 7.35x10^22 kg
Distance between Earth and Moon = 3.84x1^8

ii) Is this the point at which the combined gravitational field is zero?Explain

b)
i) Explain why an astronaut in a spacecraft orbiting the Earth appears to be weightless?
ii)State the condition necessary for true weightlessness.

Homework Equations


V=GM/r
g=Gm/r^2

The Attempt at a Solution


I think
for part a ) i)
i need to equate the potential of moon and Earth to find the distance from the Earth.
for part a) ii)
I think the answer is no . because g is inversely proportional to r^2 and not r. but this pat is 3 marks so I may need to find that point by equation the g from moon and Earth. I am not really sure.
for b)i)
I think g is nearly zero in space so resultant force is zero. but this also doesn't make sense as there must be some forces as the astronaut is orbiting the Earth.
plus this is also 3 marks.
ii)I think weightlessness is when g is nearly zero
 
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For part a) you need to do some equations.

For part b), what does "apparently" weightless mean? In physics, something is what it is measured to be. How would you measure the weight of an astronaut in orbit?

This also helps to say what "true" weightlessness is.
 
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PeroK said:
For part a) you need to do some equations.

For part b), what does "apparently" weightless mean? In physics, something is what it is measured to be. How would you measure the weight of an astronaut in orbit?

This also helps to say what "true" weightlessness is.
so your saying that what I have said for part a) i and ii are correct and I just need to add the equations and find those points!

for part b, i
probably this is what I need to do:

w=mv^2/r

as weight is the only force acting on the astronaut.
w=mg so by canceling the g's I'll get
g=v^2/r

but r is very big so g will be zero.
but I still don't know how to explain the apparent weightlessness and the true weightlessness
 
shk said:
so your saying that what I have said for part a) i and ii are correct and I just need to add the equations and find those points!

for part b, i
probably this is what I need to do:

w=mv^2/r

as weight is the only force acting on the astronaut.
w=mg so by canceling the g's I'll get
g=v^2/r

but r is very big so g will be zero.

##v## might be big as well!
 
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shk said:
true weightlessness
True weightlessness, with no other forces present, would produce what acceleration?
 
PeroK said:
##v## might be big as well!
you are right.
so maybe I should put it this way:
g is less than 9.8 in space but surely is not still zero so the astronaut still has weight and the weight is the centripetal force . this explains the speed.
but because the person is not pushing on anything so there won't be a reaction on him so he feels weightless. like in a lift.
and the true weightless is probably at that point between Moon and Earth where the g's cancel each other out !
 
haruspex said:
True weightlessness, with no other forces present, would produce what acceleration?
zero?!
 
shk said:
zero?!
Right... and the acceleration of the astronaut is...?
 
haruspex said:
Right... and the acceleration of the astronaut is...?
I think the downward acceleration is g !
 
  • #10
shk said:
I think the downward acceleration is g !
Little g is conventionally used for (average) gravitational acceleration at Earth's surface.

As for "downward", probably better to say "Earthward". My favourite Isaac Newton cartoon has him sitting under a tree contemplating an apple on the ground and observing "It fell down." The caption: Newton discovers tautology.
 
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  • #11
haruspex said:
Little g is conventionally used for (average) gravitational acceleration at Earth's surface.

As for "downward", probably better to say "Earthward". My favourite Isaac Newton cartoon has him sitting under a tree contemplating an apple on the ground and observing "It fell down." The caption: Newton discovers tautology.

I understand. I'd better say Earthward ;) thanks
But I still don't know what is the answer to this question. I need it for tomorrow. Can you please help me with this?
Thanks
 
  • #12
shk said:
I understand. I'd better say Earthward ;) thanks
But I still don't know what is the answer to this question. I need it for tomorrow. Can you please help me with this?
Thanks
Think about how you are aware of your own weight. What force do you feel?
 
  • #13
PeroK said:
##v## might be big as well!
sorry but I need the answer to this question. Do you by any chance know the answer to help me with this. many thanks
 
  • #14
shk said:
sorry but I need the answer to this question. Do you by any chance know the answer to help me with this. many thanks
Please try to answer post #12.
 
  • #15
haruspex said:
Please try to answer post #12.
contact force/ normal reaction force
 
  • #16
shk said:
contact force/ normal reaction force
Right! So apparent weightlessness consists of feeling no applied forces, so the only applied force is gravity. Real weightlessness consists of there being no gravity.
 
  • #17
haruspex said:
Right! So apparent weightlessness consists of feeling no applied forces, so the only applied force is gravity. Real weightlessness consists of there being no gravity.
so you're saying that apparent weightlessness is when normal reaction force is zero. like in free fall. when the only force acting on the object is it's weight . but the true weightless only happens when g=0 ?
 
  • #18
shk said:
so you're saying that apparent weightlessness is when normal reaction force is zero. like in free fall. when the only force acting on the object is it's weight . but the true weightless only happens when g=0 ?
Yes.
 
  • #19
haruspex said:
Yes.
so the astronaut's feeling weightless because he is in free fall meaning there is no reaction force (N=0) .
and the condition of true weightlessness is when g=0.
fir the second one can i say this point can be somewhere between Moon and Earth where g's cancel each other out like we do for part a?
although i am not if it is ok to ignore the g of sun
 
  • #20
shk said:
fir the second one can i say this point can be somewhere between Moon and Earth where g's cancel
Do you mean for b ii)?
You are only asked to state the condition. You are not asked for any specific example of where such a condition could arise, though you would not lose marks for adding that.
 
  • #21
Isn't that L1? or am I getting my numbering order wrong on the moon's LaGrange points?
 
  • #22
Ravensong said:
Isn't that L1? or am I getting my numbering order wrong on the moon's LaGrange points?
No. It is a common misunderstanding that a Lagrange point is where gravitational fields cancel.
Rather, they are points where the net field leads to the same orbital period as the larger pair of bodies.
E.g. for the Earth and Sun as the major bodies, a Lagrange point is where a much smaller body would stay at constant distances from those two.
 

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