Gravitational Force: F=G\frac{mM}{r^2} & M'=\frac{Mr^3}{R^3}

[aloshi]

Gravitational force on mass m outside a sphere with mass M is given by [tex]F=G\frac{mM}{r^2}[/tex], where r is the distance to the center of mass. Gravity inside the sphere surface because of the mass distribution, since only the portion of the sphere mass is inside r contributes to the attraction. If the Earth's density is constant (which it definitely is not), so given the mass inside r of [tex]M'=\frac{Mr^3}{R^3} [/tex], where R is Earth radius. Gravitational force in the Earth's surface (r less than R) thus becomes: [tex]F=G\frac{mMr}{R^3}[/tex]

my question is how can we prove/shown formula [tex]M'=\frac{Mr^3}{R^3} [/tex]??

 

[Doc Al]

Hint: What fraction of the total spherical volume does the mass < r occupy?

 

[aloshi]

Hint: What fraction of the total spherical volume does the mass < r occupy?

what does mean with occupy?

 

[Doc Al]

what does mean with occupy?

Occupy means 'take up', but I'll restate it differently. Since you assume uniform density, the mass is proportional to the volume. Compare the volume of a sphere of radius = r to one of radius = R.

 

[aloshi]

Occupy means 'take up', but I'll restate it differently. Since you assume uniform density, the mass is proportional to the volume. Compare the volume of a sphere of radius = r to one of radius = R.

I can not really understand how to get
[tex]M’= \frac{Mr^3 }{R^3}[/tex]
“the mass is proportional to the volume” = [tex]M(V)=\rho \frac{4\pi r^2}{3}[/tex]
if we write [tex]M(V)= \frac{\rho4\pi }{3}\cdot r^2[/tex]
[tex] \frac{\rho4\pi }{3}=k[/tex]
k=constant
[tex]M(r)=k\cdot r^2[/tex]

can you show me mathematical how I can get [tex]M’= \frac{Mr^3 }{R^3}[/tex]

 

[Doc Al]

if we write [tex]M(V)= \frac{\rho4\pi }{3}\cdot r^2[/tex]

That should be:

[tex]M(V)= \frac{\rho4\pi }{3}\cdot r^3[/tex]

Compare the total mass M_R (where radius = R) to the partial mass M_r (where radius = r).

 

[aloshi]

That should be:

[tex]M(V)= \frac{\rho4\pi }{3}\cdot r^3[/tex]

Compare the total mass M_R (where radius = R) to the partial mass M_r (where radius = r).

but there is no evidence that the mass is [tex]M'=\frac{Mr^3}{R^3}[/tex]
can you show me mathematical, so I can understand

 

[Doc Al]

but there is no evidence that the mass is [tex]M'=\frac{Mr^3}{R^3}[/tex]
can you show me mathematical, so I can understand

What does M equal? (Use the formula for density times volume.)
What does M' equal? (Use the formula for density times volume.)

Then just divide M' by M and see what you get.

 

[aloshi]

What does M equal? (Use the formula for density times volume.)
What does M' equal? (Use the formula for density times volume.)

Then just divide M' by M and see what you get.

[tex]M=\rho \frac{4\pi \cdot R^3}{3}\\ [/tex]
[tex]M'=\rho \frac{4\pi r^3}{3}[/tex]
[tex]\frac{M'}{M}=\frac{\rho \frac{4\pi r^3}{3}}{\rho \frac{4\pi \cdot R^3}{3}}[/tex]

this does not give us the mass, this give is the share

 

[Doc Al]

Finish the division--canceling things that can be canceled--and you'll get the formula you want.

 

[aloshi]

Finish the division--canceling things that can be canceled--and you'll get the formula you want.

[tex]\frac{M'}{M}=\frac{r^3}{R^3}[/tex]
but that's not what I want, this give me the share, not the new mass

 

[Doc Al]

[tex]\frac{M'}{M}=\frac{r^3}{R^3}[/tex]
but that's not what I want, this give me the share, not the new mass

Multiply both sides by M.