Gravitational Force Between; Sun and Earth, Moon and Earthby NewtonJR.215 Tags: earth, force, gravitational, moon 

#1
May1508, 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 



#2
May1508, 05:23 PM

Mentor
P: 40,878

[tex]F = \frac{G M_1 M_2}{R^2}[/tex] (Look up Newton's law of universal gravity.) 



#3
May1508, 05:28 PM

P: 10

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. 



#4
May1508, 05:34 PM

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P: 40,878

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.




#5
May1508, 05:39 PM

P: 10

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. 



#6
May1508, 05:53 PM

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P: 40,878





#7
May1508, 06:15 PM

P: 10

I made it a much smoother number. Thanks for the help Doc.
This Problem is SOLVED. 



#8
May1508, 06:26 PM

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P: 40,878

Excellent. Mark it Solved: How to mark problems SOLVED




#9
Mar1209, 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. 



#10
Mar1209, 07:30 AM

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P: 40,878





#11
Mar1209, 08:10 AM

P: 6

thank you very much, this clears up a lot for me




#12
May109, 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 



#13
May109, 07:00 PM

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P: 40,878

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!) 



#14
Jul1909, 11:31 PM

P: 1

I came up with 3.5428*10^22N. Other sites have answers that approximate this value.




#15
Aug409, 05:50 PM

P: 2





#16
Aug409, 09:18 PM

HW Helper
P: 2,324





#17
Aug409, 09:21 PM

HW Helper
P: 2,324





#18
Aug409, 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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