Earth's gravity on the Moon

Earth's gravity on the Moon

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Derek Smith

Original Poster:

45,514 posts

247 months

Sunday 15th July 2018
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I've tried Google but can't get what I want to win a bet.

What is the % of G on an object in the Moon's orbit.


Eric Mc

121,788 posts

264 months

Sunday 15th July 2018
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It would be a fraction ofa percentage but I wouldn't know how much.

At what distance from the moon is your hypothetical object orbiting?

S6PNJ

5,158 posts

280 months

Sunday 15th July 2018
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Just checking...
Do you mean the % of gravity (9.81m/s/s) or big G?

(copied and pasted) The English polymath knew the objects' masses had to be multiplied by a constant, or "big G," in order to arrive at the gravitational force between those two objects, but he wasn't able to calculate its value. ("Big G" is different from "little g," which is the local gravitational acceleration on Earth.

FarmyardPants

4,099 posts

217 months

Sunday 15th July 2018
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Derek Smith said:
I've tried Google but can't get what I want to win a bet.

What is the % of G on an object in the Moon's orbit.
Do you actually mean “in orbit” (around the Earth, like the moon) because any object orbiting a mass will not experience any effects of that mass’s gravitational field (it will experience weightlessness) and you could argue that the answer is zero.

If you mean “what fraction of Earth’s gravity, relative to that experienced on the Earth’s surface, is experienced by an object at a distance equal to the distance between the earth and moon” then it would be a small fraction but obviously > 0.


Edited by FarmyardPants on Sunday 15th July 18:22

JonChalk

6,469 posts

109 months

Sunday 15th July 2018
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Do you mean what effect caused by the Earth would a mass "feel" on the moon?

Or, if the mass was on the moon, with Earth directly above, how much "pull" would that mass feel from Earth?


Derek Smith

Original Poster:

45,514 posts

247 months

Sunday 15th July 2018
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FarmyardPants said:
If you mean “what fraction of Earth’s gravity, relative to that experienced on the Earth’s surface, is experienced by an object at a distance equal to the distance between the earth and moon” then it would be a small fraction but obviously > 0.
That's the one.

The conversation, before the match, was around objects in a NEO. Then it went onto when it would 'run out.'

My contention is that it would be significant 250m miles away, i.e. a little over the Moon's orbit. Significant meaning that it would be still under the control of the Earth's gravity. I the scenario we were discussing, being hit by the Moon, or even having it miss slightly, was not an option.


Eric Mc

121,788 posts

264 months

Sunday 15th July 2018
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Objects out as far as the moon can still be in orbit around the earth. i.e. under the control of earth's gravity. Indeed, just as the moon itself is.

Kccv23highliftcam

1,783 posts

74 months

Sunday 15th July 2018
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This is a good one indeed.

As Farmyard Pants [btw brilliant handle] and Eric rightly pointed out [in plain English] the moon is already captured by Earths gravitational force, so any gravitational effect on an object in lunar orbit is under the earths effect.

However It must also be balanced under the moons, as otherwise it would exit stage left, rapido...


SCEtoAUX

4,119 posts

80 months

Sunday 15th July 2018
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My maths says this:

https://www.mansfieldct.org/Schools/MMS/staff/hand...

Stick 384,000 into that equation and you're looking at about 1/4000th of the Earth's gravity when at an average lunar distance. Insignificant compared to that which the moon would exert if you were stood on it.

Derek Smith

Original Poster:

45,514 posts

247 months

Sunday 15th July 2018
quotequote all
Kccv23highliftcam said:
This is a good one indeed.

As Farmyard Pants [btw brilliant handle] and Eric rightly pointed out [in plain English] the moon is already captured by Earths gravitational force, so any gravitational effect on an object in lunar orbit is under the earths effect.

However It must also be balanced under the moons, as otherwise it would exit stage left, rapido...
The fact that the Moon orbits the Earth, or rather the Moon and the Earth orbit their common centre of gravity, is what started the argument.

This is what I was after:

SCEtoAUX said:
My maths says this:

https://www.mansfieldct.org/Schools/MMS/staff/hand...

Stick 384,000 into that equation and you're looking at about 1/4000th of the Earth's gravity when at an average lunar distance. Insignificant compared to that which the moon would exert if you were stood on it.
Thank you for that.

By the way, in my Googling I discovered that an object in a near Earth orbit would have circa 9/10ths of full gravity.

No help to me, but interesting.


James_B

12,642 posts

256 months

Monday 16th July 2018
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Gravity drops off with the square of the distance, so that’s the only thing you need for this, the ratio of the earth’s radius to the distance to the moon.

annodomini2

6,860 posts

250 months

Monday 16th July 2018
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If you were able to instantly travel ~4Bn ly from Earth, you would still have some influence of Earth's gravity, albeit tiny.

Gravitational effect propagates at the speed of light, so over sufficient time Earth's gravity could have a very tiny effect on objects on the other side of the Universe.

Simpo Two

85,151 posts

264 months

Tuesday 17th July 2018
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James_B said:
Gravity drops off with the square of the distance, so that’s the only thing you need for this, the ratio of the earth’s radius to the distance to the moon.
So light and gravity have something in common... I just need to come up with a killer equation to get my Nobel Prize...

prand

5,910 posts

195 months

Tuesday 17th July 2018
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SCEtoAUX said:
Stick 384,000 into that equation and you're looking at about 1/4000th of the Earth's gravity when at an average lunar distance. Insignificant compared to that which the moon would exert if you were stood on it.
Ok I may be getting this all wrong, but surely the earth must exert some significant gravitational effect on the moon, not just to capture it in orbit?

My reasoning is that the moon is much smaller than earth, but still creates the tides, affects plants, and animals (and large buildings?). Plus the earth does cause the moon to be egg shaped.

Eric Mc

121,788 posts

264 months

Tuesday 17th July 2018
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prand said:
Ok I may be getting this all wrong, but surely the earth must exert some significant gravitational effect on the moon, not just to capture it in orbit?

My reasoning is that the moon is much smaller than earth, but still creates the tides, affects plants, and animals (and large buildings?). Plus the earth does cause the moon to be egg shaped.
The earth's gravitational pull on the moon does indeed create tides on the moon. Of course, there is no water (or any other liquid) on the moon's surface so the tidal effects are felt on the moon's crust. The moon flexes as it passes around the earth every 28 days. The seismometers left behind by the Apollo missions registered an increase in moonquakes at the times of "high tide" on the moon.
It is the earth's gravity that has slowed the moon's rotational rate down to the point where one face always points towards the earth.

The moon is doing the same to the earth only at a slower rate - because it is a much smaller body.


FarmyardPants

4,099 posts

217 months

Tuesday 17th July 2018
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But since the moon is “tidally locked”, it is permanently high tide, as it were, on the side facing the Earth (and also the opposite side). That said, the moon is not in a circular orbit so the effect of Earth’s gravity varies causing the bulging of the moon to vary slightly during the month (and also Earth’s tides to vary of course).

SCEtoAUX

4,119 posts

80 months

Tuesday 17th July 2018
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prand said:
SCEtoAUX said:
Stick 384,000 into that equation and you're looking at about 1/4000th of the Earth's gravity when at an average lunar distance. Insignificant compared to that which the moon would exert if you were stood on it.
Ok I may be getting this all wrong, but surely the earth must exert some significant gravitational effect on the moon, not just to capture it in orbit?

My reasoning is that the moon is much smaller than earth, but still creates the tides, affects plants, and animals (and large buildings?). Plus the earth does cause the moon to be egg shaped.
Sure it does, but if the moon were right next to the earth, the gravity would be 4,000 times greater.

With regards to the 9/10ths mentioned above, it should be noted that the guys in the space station experience this. They weigh 9/10ths of what they do on earth, and their mass is the same. It's just that the constant "falling" around the earth gives the effect of weightlessness.

Eric Mc

121,788 posts

264 months

Tuesday 17th July 2018
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FarmyardPants said:
But since the moon is “tidally locked”, it is permanently high tide, as it were, on the side facing the Earth (and also the opposite side). That said, the moon is not in a circular orbit so the effect of Earth’s gravity varies causing the bulging of the moon to vary slightly during the month (and also Earth’s tides to vary of course).
And there is also the influence of the sun - which also causes smaller tides.

James_B

12,642 posts

256 months

Tuesday 17th July 2018
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Simpo Two said:
So light and gravity have something in common... I just need to come up with a killer equation to get my Nobel Prize...
Yes, because they are both mediated by massless bosons, so far as we can tell.

prand

5,910 posts

195 months

Wednesday 18th July 2018
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FarmyardPants said:
But since the moon is “tidally locked”, it is permanently high tide, as it were, on the side facing the Earth (and also the opposite side). That said, the moon is not in a circular orbit so the effect of Earth’s gravity varies causing the bulging of the moon to vary slightly during the month (and also Earth’s tides to vary of course).
Ah I see, the moon isn't spinning any more which means the motion effect from tides is not really happening like it does on earth, so the effects are not so pronounced, not like that moon around Jupiter made up of a liquid interior that continually erupts through the moon's crust due to the combination of gravity and movement around the gas giant.