Question

pls solve it with steps n appropriate reasons....​

Answer 1

Ans -7/8)mgR

I’m going to work with gravitational acceleration and potential for most of the calculation, simply to avoid redundantly including body mass m in all the equations. Let’s say that we have the following basic information about gravity due to such an object as given:

g(r)=GMr2

V(r)=−GMr

when r>R , where g(r) and V(r) are gravitational acceleration and potential due to the object at distance r from the centre. We’ll also take the Shell theorem[1] as a given, and that gravitational acceleration is the distance derivative of the potential (equivalent to the force on a body being the distance derivative of potential energy of the body).

First let’s calculate g(r) for r<R . The Shell theorem tells us that in this case we only need to consider the mass of the ball of radius r , and the uniform density means that this mass is proportional to the volume of this ball which in turn is proportional to r3 . When r=R then the mass is M , and therefore the mass of the ball radius r is:

M(r)=r3R3M

Plugging this into the equation for g(r) we get:

g(r)=Gr3R3Mr2

=GMrR3

for r<R .

Now as this is the derivative with respect to r of potential, then to get the potential at distance r<R from the centre we need to integrate with respect to r :

V(r)=∫g(r) dr=∫GMrR3 dr

=GMr22R3+C

for some integration constant C . We know V(R) and using this we have:

V(R)=GMR22R3+C=−GMR

Solving for C :

C=−3GM2R

giving:

V(R)=GM2R(r2R2−3)

We’re asked for the work done from r=½R to r=2R , so we need the potential difference between these two distances. Using the respective equations we get:

V(2R)−V(½R)=−GM2R−GM2R(¼R2R2−3)

=GM2R(3−¼−1)=78GMR

Noting that gR=Rg(R)=GMR , and multiplying potential difference by the body mass m to get the work done on the body, we end up with the expected 78mgR .

IF YOU LIKE THIS ANSWER

SO PLEASE MARK BRILLIANT ANSWER