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 $$F=G\frac{mM}{r^2}$$, 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 $$M'=\frac{Mr^3}{R^3}$$, where R is Earth radius. Gravitational force in the Earth's surface (r less than R) thus becomes: $$F=G\frac{mMr}{R^3}$$

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

 

[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
$$M’= \frac{Mr^3 }{R^3}$$
“the mass is proportional to the volume” = $$M(V)=\rho \frac{4\pi r^2}{3}$$
if we write $$M(V)= \frac{\rho4\pi }{3}\cdot r^2$$
$$\frac{\rho4\pi }{3}=k$$
k=constant
$$M(r)=k\cdot r^2$$

can you show me mathematical how I can get $$M’= \frac{Mr^3 }{R^3}$$

 

[Doc Al]

if we write $$M(V)= \frac{\rho4\pi }{3}\cdot r^2$$

That should be:

$$M(V)= \frac{\rho4\pi }{3}\cdot r^3$$

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

 

[aloshi]

That should be:

$$M(V)= \frac{\rho4\pi }{3}\cdot r^3$$

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 $$M'=\frac{Mr^3}{R^3}$$
can you show me mathematical, so I can understand

 

[Doc Al]

but there is no evidence that the mass is $$M'=\frac{Mr^3}{R^3}$$
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.

$$M=\rho \frac{4\pi \cdot R^3}{3}\\$$
$$M'=\rho \frac{4\pi r^3}{3}$$
$$\frac{M'}{M}=\frac{\rho \frac{4\pi r^3}{3}}{\rho \frac{4\pi \cdot R^3}{3}}$$

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.

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

 

[Doc Al]

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

Multiply both sides by M.