The gravitational field g due to a point

[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.

-------------------------------------------

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

 

[voko]

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

 

[BadSkittles]

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

 

[voko]

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

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

 

[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

 

[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?

 

[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?

 

[voko]

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

Second, what is $M$ in this case?

 

[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.

 

[voko]

∫ dA is your gaussian surface.

And what is the surface here?

M is the Mass of the whole planet.

Even including that outside the surface?

 

[BadSkittles]

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

 

[voko]

So the gaussian surface is 4 pi r^2 still

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

but the mass is a ratio between V and m?

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

 

[BadSkittles]

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

 

[voko]

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