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Gravitational Force Between; Sun and Earth, Moon and Earth

by NewtonJR.215
Tags: earth, force, gravitational, moon
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NewtonJR.215
#1
May15-08, 05:09 PM
P: 10
1. The problem statement, all variables and given/known data
"Calculate the gravitational force between:
A) The sun and the earth
B)The moon and the earth"


2. Relevant equations
F= M/r2


3. The attempt at a solution
A)The sun and the earth
Earth: mass 5.97*10^24 kg
Radius 6,380,000 meters
Sun: mass 1.99*10^30
Radius 696,000,000
Distance Between Sun &Earth: 1.5*10^11 meters (149,668,992,000 meters)

(5.97*10^24)/1.5*10^11

=3.98*10^13


B)
The moon and the earth
Moon: mass 7.35 *10^22 kg
Radius 1,740,000 meters
Earth: mass 5.97*10^24 kg
Radius: 6,380,000 meters
Distance between moon and earth: 384,000,000 meters

5.97*10^24/384,000,000
= 1.55*10^16

**I'm not sure if i'm doing this correctly. I'm not sure if I should use the mass of the sun/moon also or just the earth's mass. Also not sure if i should use the distance between the planets...

Thanks for looking it over!
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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Doc Al
#2
May15-08, 05:23 PM
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Quote Quote by NewtonJR.215 View Post
2. Relevant equations
F= M/r2
The gravitational force between two astronomical bodies depends on both their masses and the distance between them:

[tex]F = \frac{G M_1 M_2}{R^2}[/tex]

(Look up Newton's law of universal gravity.)
NewtonJR.215
#3
May15-08, 05:28 PM
P: 10
Quote Quote by Doc Al View Post
The gravitational force between two planets (or planet and moon) depends on both their masses and the distance between them:

[tex]F = \frac{G M_1 M_2}{R^2}[/tex]

(Look up Newton's law of universal gravity.)
So for sun and earth... 6.67*10^-11(5.97*10^24)(1.99*10^30)/ r2

I'm not sure what the radius would be. I thought it would be the distance between the sun and the earth. Which radius would I use, that of the earth or that of the sun?


For the moon and earth...6.67*10^-11(5.97*10^24)(7.35*10^22)/r2

Once again, not sure which radius I would use.

Doc Al
#4
May15-08, 05:34 PM
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Gravitational Force Between; Sun and Earth, Moon and Earth

The "R" in the formula for gravity that I gave stands for the distance between the two bodies, not the radius.
NewtonJR.215
#5
May15-08, 05:39 PM
P: 10
Quote Quote by Doc Al View Post
The "R" in the formula for gravity that I gave stands for the distance between the two bodies, not the radius.
Oh. Thanks for the response.

So for sun and earth use: 1.5*10^11^2 .... or 149,668,992,000^2


and for moon and earth use: 384,000,000^2

....? thanks again doc.
Doc Al
#6
May15-08, 05:53 PM
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Quote Quote by NewtonJR.215 View Post
Oh. Thanks for the response.

So for sun and earth use: 1.5*10^11^2 .... or 149,668,992,000^2
It doesn't much matter, but 9 significant figures is overkill. But let the calculator do the work. Generally, I advise you to use the most accurate numbers given and only round off your final answer.
and for moon and earth use: 384,000,000^2
That's fine.
NewtonJR.215
#7
May15-08, 06:15 PM
P: 10
I made it a much smoother number. Thanks for the help Doc.

This Problem is SOLVED.
Doc Al
#8
May15-08, 06:26 PM
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Excellent. Mark it Solved: How to mark problems SOLVED
billyx3
#9
Mar12-09, 06:47 AM
P: 6
Sorry for posting on a nearly year old thread and I'm not expecting responses from the original posters but for anybody out there, I've got a question I'm curious about.
I googled "gravitational force earth moon" and all I could find is this thread, I was actually looking for the exact figures for the force. Anyways, I thought in the equation (Gm1m2)/(R^2), the R is the distance between the centers of the masses. In other words, the radius of the moon, plus the radius of the earth, plus the average distance between the moon and earth, so not just the average distance between the moon and earth alone. Please correct me if I'm wrong because it will affect my numbers.
Doc Al
#10
Mar12-09, 07:30 AM
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Quote Quote by billyx3 View Post
In other words, the radius of the moon, plus the radius of the earth, plus the average distance between the moon and earth, so not just the average distance between the moon and earth alone.
Generally, when figures are given for the average distance between moon and earth they are talking about center to center distance. For example, see: Wiki: Moon.
billyx3
#11
Mar12-09, 08:10 AM
P: 6
thank you very much, this clears up a lot for me
kootzie
#12
May1-09, 06:27 PM
P: 1
I came across this site
while following up a funky proposition I found here:
<crackpot link deleted>

which states that the sun's pull on the moon, during a new moon,
is almost 3X that of the earth
and asks why the moon doesn't just take off

When I pound the numbers through G*m1*m2/dē
I get min/max numbers of 1.77E20 to 2.3E20 N for Earth<->Moon
and min/max numbers of 4.42E20 to 4.49E20 N for Sun<->Moon

I've checked the numbers a few times, and used the calculator at
http://www.ajdesigner.com/phpgravity...tion_force.php
and get the same results.

Either I'm making some mistake in the math, or
What Ralph Rene claims is a very interesting question...

kk
Doc Al
#13
May1-09, 07:00 PM
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Quote Quote by kootzie View Post
I came across this site
while following up a funky proposition I found here:
<crackpot link deleted>

which states that the sun's pull on the moon, during a new moon,
is almost 3X that of the earth
and asks why the moon doesn't just take off
Careful what you read on crackpot sites. I didn't check your calculations, but the force of the sun on the moon is greater than the earth on the moon. So what? Both moon and earth are in orbit around the sun.

A more interesting question would be: What's the difference in the strength of the sun's gravitational field at the moon's various locations versus the earth's location. (Answer: Not much!)
RonArt
#14
Jul19-09, 11:31 PM
P: 1
I came up with 3.5428*10^22N. Other sites have answers that approximate this value.
koko86
#15
Aug4-09, 05:50 PM
P: 2
Quote Quote by Doc Al View Post
A more interesting question would be: What's the difference in the strength of the sun's gravitational field at the moon's various locations versus the earth's location. (Answer: Not much!)
Good day new poster here, I was wondering how we would calculate this question? Well my question specifically would be can we calculate the gravitational forces exerted by the moon to a point on the Earth on an hourly basis?
ideasrule
#16
Aug4-09, 09:18 PM
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Quote Quote by kootzie View Post
I came across this site
while following up a funky proposition I found here:
<crackpot link deleted>

which states that the sun's pull on the moon, during a new moon,
is almost 3X that of the earth
and asks why the moon doesn't just take off

When I pound the numbers through G*m1*m2/dē
I get min/max numbers of 1.77E20 to 2.3E20 N for Earth<->Moon
and min/max numbers of 4.42E20 to 4.49E20 N for Sun<->Moon

I've checked the numbers a few times, and used the calculator at
http://www.ajdesigner.com/phpgravity...tion_force.php
and get the same results.

Either I'm making some mistake in the math, or
What Ralph Rene claims is a very interesting question...

kk
Think about an astronaut in the ISS. Earth exerts a much greater force than the ISS on the astronaut, but he doesn't fall to the Earth. Why? Because both he and the ISS ARE falling towards the Earth; they're both in orbit, so the Earth imparts the same acceleration to both objects.
ideasrule
#17
Aug4-09, 09:21 PM
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P: 2,322
Quote Quote by koko86 View Post
Good day new poster here, I was wondering how we would calculate this question? Well my question specifically would be can we calculate the gravitational forces exerted by the moon to a point on the Earth on an hourly basis?
What do you mean "a point on the Earth"? Gravity only acts on objects with mass; a geometrical point is not a massive object. That said, you can certainly calculate the Moon's gravitational field. Currently-available astronomical data is more than precise enough to reveal hour-by-hour differences.
koko86
#18
Aug4-09, 11:51 PM
P: 2
By point I mean would the gravitational force being exerted by the Moon on a body of water say in Sydney (Australia) be slightly different to the amount being exerted in say Tokyo Japan, assuming they share the same timezone but have very different coordinates.


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