# The gravitational field g due to a point

1. Jul 16, 2013

### BadSkittles

The gravitational field g due to a point mass M may be obtained by analogy with the electric field by writing an expression for the gravitational force on a test mass, and dividing by the magnitude of the test mass, m. Show that Gauss' law for the gravitational field reads:

phi = oint g*dA=-4*pi*GM

where G is the gravitational constant.

Use this result to calculate the gravitational acceleration g at a distance of R/2 from the center of a planet of radius R = 6.15 x 1006 m and M = 4.25 x 1024 kg.

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Hi, I got the final equation to be g= G M / r^2. My final result was -29.9 m/s^2. But thats not correct. I dont see what i did wrong. Anyone has any ideas?

2. Jul 16, 2013

### voko

That equation is correct outside the planet. But you need the acceleration at R/2 from the center.

3. Jul 16, 2013

### BadSkittles

does that mean i would have to take a fraction of the mass, because of the radius?

4. Jul 16, 2013

### voko

It probably does, but then you need to know the distribution of mass inside the planet.

5. Jul 16, 2013

### BadSkittles

I tried to divide the mass by 2, and the answer is still wrong. Do you know any way to solve this problem T-T

6. Jul 16, 2013

### voko

Why would you divide the mass by 2, and not by 123, for example? How did you use Gauss's law to obtain the equation?

7. Jul 16, 2013

### BadSkittles

I got g * Integral of dA = -4*pi G*M

g * 4 * pi * r^2= -4* pi* G*M

g= - GM/ r^2

Maybe the integral of dA is something else?

Last edited: Jul 16, 2013
8. Jul 16, 2013

### voko

First of all, what is $\int dA$ if you are asked for $g$ at $R/2$?

Second, what is $M$ in this case?

9. Jul 16, 2013

### BadSkittles

∫ dA is your gaussian surface. Would that be 4/3 pi r^3 ? Since our radius is shortened. M is the Mass of the whole planet.

10. Jul 16, 2013

### voko

And what is the surface here?

Even including that outside the surface?

11. Jul 16, 2013

### BadSkittles

So the gaussian surface is 4 pi r^2 still, but the mass is a ratio between V and m?

Last edited: Jul 16, 2013
12. Jul 16, 2013

### voko

You did not answer the question. Explain the shape of the surface, and then what its area is.

Assuming V is volume and m is mass, how can mass be a ratio between volume and mass? That is dimensionally impossible.

13. Jul 16, 2013

### BadSkittles

I think its m = M ( r^3/ R^3)

14. Jul 16, 2013

### voko

R is the radius of the planet, and M is its mass. What are r and m?

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