# Another gravitation problem

1. Sep 7, 2016

### Deebu R

1. The problem statement, all variables and given/known data
If the gravitational accleration at the Earths surface is 9.81 m/s^2 what is its value at a height equal to the diameter of the earth from its surface?

2. The attempt at a solution
I have heard that it becomes 1/4th the value as the center of mass is moving away form each other.
But I really can't understand the idea. Why is it 1/4?

Last edited: Sep 7, 2016
2. Sep 7, 2016

### PeroK

What do you know about the force gravity?

3. Sep 7, 2016

### Deebu R

I know that it is a force acting between 2 masses with a distence between them. It is represented by F= G m1m2/ d^2.
I also know that it is a concervative force and a central force.

4. Sep 7, 2016

### PeroK

That's good. So, what can you do with that formula for $F$?

5. Sep 7, 2016

### Deebu R

Well I don't understand how I can apply it here. How can we use the formula of force to find gravitational accelereration? Why not find the gravitational acceleration at a hight h from surface of Earth?

6. Sep 7, 2016

### PeroK

Are force and acceleration not related?

7. Sep 7, 2016

### Deebu R

Yes but I still don't understand how that solve the problem..... Say I use the formula G=9.81m/s^2. Earths radius is 6400km so its diameter is x2. what about mass?
should it be volume x density? Won't that make the problem unnecessarily large and complicated?

8. Sep 7, 2016

### PeroK

One thing at a time! Let me help. You have a general formual for the gravitational force. You want the gravitational acceleration due to the Earth:

1) On the Earth's surface

2) At a height equal to the diameter of the Earth above its surface.

So, you need:

Let $M$ be the mass of the Earth, let $R$ be the radius of the Earth. Now, what can you can about the accleration of gravity at the Earth's surface?

9. Sep 7, 2016

### Deebu R

Acceleration due to gravity at the earths surface = GM/R^2

10. Sep 7, 2016

### PeroK

What about above the Earth's surface, at a height equal to the Earth's diameter?

11. Sep 7, 2016

### Deebu R

Well, using Gp= GM/(R+h)^2. So....9.81 x M/ (6400+12800)^2 ?
Isn't M= volume x density. How can I find M withought density?

12. Sep 7, 2016

### BvU

Work with symbols. Leave it in as M. It's the same for h = 0 as for any other value of h

13. Sep 7, 2016

### Deebu R

So...... 9.81 x M /(6400)^2+h^2?

14. Sep 7, 2016

### BvU

No. And I still see numbers, so I haven't been clear enough.
You have
1) at ground level $$g = {GM\over R^2}$$
2) at height h $$g^* = {GM\over (R+h)^2}$$All you want to do is pick a good value for h and express $g^*$ in terms of $g$

15. Sep 7, 2016

### Deebu R

Already pointed out both equations but I don't get the value part....earths radius=6400km so shouldn't its diameter be 6400 x 2.
Is it the convertion? should I make it into meter?

16. Sep 7, 2016

### Deebu R

Also what is the value of M?

17. Sep 7, 2016

### Deebu R

Ahhhh I see G is the gravitational constent 6.674x 10^11 and M = gr^2/G.Right?I though G was g.This was my own mistake.
But still can't get the right answer. Maybe I need to understand the concept more deeply.

Last edited: Sep 7, 2016
18. Sep 7, 2016

### Staff: Mentor

In BvU's post # 14, algebraically, what is g*/g equal to? Do you need to know the mass of the earth M to get this ratio?

19. Sep 7, 2016

### Deebu R

Okay. I have to find the ratio..so GM cancel out...

so g*/g= R^2/(R+h)?

20. Sep 7, 2016

### Staff: Mentor

Yes, except the R+h is also squared. Does this give you your desired answer?