# Gravitational Force Between; Sun and Earth, Moon and Earth

1. May 15, 2008

### NewtonJR.215

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
Sun: mass 1.99*10^30
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
Earth: mass 5.97*10^24 kg
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. May 15, 2008

### Staff: Mentor

The gravitational force between two astronomical bodies depends on both their masses and the distance between them:

$$F = \frac{G M_1 M_2}{R^2}$$

(Look up Newton's law of universal gravity.)

Last edited: May 15, 2008
3. May 15, 2008

### NewtonJR.215

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.

4. May 15, 2008

### Staff: Mentor

The "R" in the formula for gravity that I gave stands for the distance between the two bodies, not the radius.

5. May 15, 2008

### NewtonJR.215

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.

6. May 15, 2008

### Staff: Mentor

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.
That's fine.

7. May 15, 2008

### NewtonJR.215

I made it a much smoother number. Thanks for the help Doc.

This Problem is SOLVED.

8. May 15, 2008

### Staff: Mentor

9. Mar 12, 2009

### billyx3

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.

Last edited: Mar 12, 2009
10. Mar 12, 2009

### Staff: Mentor

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.

11. Mar 12, 2009

### billyx3

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

12. May 1, 2009

### kootzie

I came across this site
while following up a funky proposition I found here:

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/d²
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/newtons_law_gravity_equation_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

Last edited by a moderator: May 1, 2009
13. May 1, 2009

### Staff: Mentor

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!)

14. Jul 19, 2009

### RonArt

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

15. Aug 4, 2009

### koko86

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?

16. Aug 4, 2009

### ideasrule

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.

17. Aug 4, 2009

### ideasrule

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.

18. Aug 4, 2009

### koko86

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.

19. Mar 12, 2010

### pjdonkeykong

well, since the distances from the moon to australia and the moon to tokyo are different, the forces of gravity will also be different in those locations, however, for practical purposes, the difference would be negligible since it is so small.

on another note, may be a dumb question, but i thought that the r was the radius of the orbit, meaning this only applies to circular motion. but since the orbits of planets and sattelites is actually elliptical, wouldnt these values be off?

aha again, reviving a dead post =)

20. Jul 7, 2010

### Cybr

I have some considerations that come along with this topic, but might be separated into a new topic, or not?

* Can be concluded that the sun and moon gravity forces on earth are a reasonable percentage of eachother, so that both forces in one direction (midummes and midwinter new-moon), have a significant additional effect?

* what is the mentioned effect of the sun pulling on the moon at new moon, towards the total effect of the gravity force moon+sun on the earth?

* as the earth is in an orbit around the sun, the moon in an orbit around the sun and earth, there are significant other forces. Forgot the name (and my English...).

* These forces change heavily within 24 hours.

* Although the total gravity forces on earth are practically the same, the earth side the forces are pulling and ... (=word I forgot), change rapidly.

* I believe to have noticed an increase in earthquakes and magnitudes around midsummer and midwinter, especially around new(=no) moon. Except for historic statistic analysis, what could be the theory and calculation for those forces and probability?