Question

The gravitational force on an object of mass m, located inside the earth a distance r < Re, from the center (Re is the radius of the earth), is due only to the mass of the earth that lies within a solid sphere of radius r. What is the gravitational force as a function of the distance r from the center? Do not look up or use any values yet! (5 pts)

Answer 1

The gravitational force is
`\r F_(g) = (mg)/(R_e) \r rr`

Step-by-step explanation:

The sketch describing this question is shown on the first uploaded image

from the sketch
`\r r` represent a unit vector pointing outward (This is shown the distance and the direction of the object

Now the gravitational force of a object on the surface earth with mass m is

Mathematically represented as

`\u F = - (Gmm_e)/(R_e^2) =- mg \r r ----(1)`

Where
`m_e` is the mass o the earth and

`R_e` is the radius of the earth

The minus sign shows that it is acting in the negative x-axis

Making the acceleration due to gravity a subject of the formula we have

`g = (Gm_e)/(R_e^2) ----(2)`

Considering when the distance of the object to the center of the earth is r

Note: Any mass of the earth that is outside the distance covered by radius r as shown in the diagram does not affect the gravitational force

The mass of the enclosed earth can be mathematically represented as

`M_(en) = \rho (4)/(3) \pi r^3 ----(3)`

Where
`\rho` is the mass density which is mathematically represented as

`\rho = (m_e)/(((4)/(3) )\pi R_e^3) ----(4)`

Now substituting these equation 3

`M_(en) = (,m_e)/(((4)/(3) )\pi R_e^3) ((4)/(3) ) \pi r^3 = (m_er^3)/(R_e^3) ---(5)`

Now substituting equation 5 into equation 1 to obtain the gravitational force as a function of distance r

`\r F = -(GmM_(en))/(r^2) \r r = (Gmm_e r^3)/(r^2R_e^3) \r r = (Gmm_e)/(R_e^3) r ----(6)`

From equation 2 we see that the
`g = (Gm_e)/(R_e^2)`

So substituting for g we have

`\r F_(g) = (mg)/(R_e) \r rr`