Two particles, each of mass m, are initially at rest very far apart. Obtain an expression for their relative speed of approach at any instant as a function of their separation distance d if the only interaction is their gravitational attraction to each other. Express your answer in terms of some or all of the variables m, d, and gravitational constant G.

Explanation:

Using the following law

Laws of Motion,

Law of Universal Gravitation, Conservation of Momentum and Energy.

The gravitation potential should be equal to the kinetic energy gain by the two bodies

Given that

The bodies have equal mass i.e M1=M2=m

Let the distance between the two masses be "d".

The gravitational potential energy is given as

P.E = force of attraction between the two bodies × distance travel.

P.E= GM1M2/r² × d

Then, M1=M2=m,, r=d

P.E=Gm²/d² × d

P.E=Gm²/d

Now the kinetic energy of the two masses

We know that K.E is given as

KE= ½MV²

So, K.E of the bodies is

K.E= ½M1•V1²+ ½M2•V2²

Also conservation of momentum

M1V1=M2V2

Since M1=M2=m

mV1=mV2

Then, divide through by m

Then, V1=V2

So the two bodies are moving with the same velocity

Then, we can say V1=V2=v

So back to our kinetic energy

K.E= ½M1•V1²+ ½M2•V2²

K.E=½mv²+½mv²

K.E=mv²

Using conversation of energy

So the P.E is equal to K.E

P.E= K.E

Gm²/d=mv²

Divide both side by m

Gm/d=v²

Take square root of both sides

v=√(Gm/d)

So, the velocity of the two bodies is a function of G, m and d

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.