Question

An experiment is performed in deep space with two uniform spheres, one with mass 20.0 kg and the other with mass 108.0 kg . They have equal radii, r = 0.30 m . The spheres are released from rest with their centers a distance 40.0 m apart. They accelerate toward each other because of their mutual gravitational attraction. You can ignore all gravitational forces other than that between the two spheres.

A. When their centers are a distance 28.0 m apart, find the speed of the 20.0 kg sphere.

B. Find the speed of the sphere with mass 108.0 kg

C. Find the magnitude of the relative velocity with which one sphere is approaching to the other.

D. How far from the initial position of the center of the 20.0 kg sphere do the surfaces of the two spheres collide?

Answer 1

A). Gravitaional potential energy at separation r is given by U= -(Gm1m2)/r (G=universal gravitational constant).
So, change in potential energy in coming close to 28m from 40 = -G*m1*m2*(1/40 -1/28)=
= 0.0107*G*m1*m2
Now, this much potential energy change will appear as their kinetic energy. Hence,
1/2m1v1^2 + 1/2m2v2^2 = 0.0107*G*m1*m2 ...(1)
As there is no external force, so momentum of the system is conserved, m1v1= m2v2
or, v2= m1v1/m2 ....(2)
from (1) and (2)

m1v1^2 + (m1^2*v1^2)/m2 = 0.001*G*m1*m2
or, v1^2*(1+ m1/m2) = 0.001*G*m2

or v1^2(1+30/108) = 0.001*G*108
or v1^2 = (0.001*G*108*108)/128 = 0.091G m/s = 6.67*0.091*10^-11 m/s = 6.069*10^-12 m/s

B). Speed of 108 kg sphere= v2= m1v1/m2= (30*)/6.069*10^-12)108 m/s = 4.576*10^-10m/s

C). magnitude of relative velocity = v1+v2 = ( 6.069*10^-12+ 4.576*10^-10) m/s = 4.63669*10^-10 m/s

D). As you can see v2/v1= m1/m2. Hence at any instant distance travelled by them will also be in inverse ratio of their masses. so d1/d2= m2/m1 .....(1)

when their surfaces will collide let center of 20kg sphere will travel say D1 distance and 108 kg will travel D1 then (D1+0.20)+(D2+0.20)= 40 or D2= 39.6-D1 ......(2)
from (1) and (2)
D1/(39.6-D1)= 108/20
or D1= 213.84 - 5.4D1
or D1= 213.84/6.4 = 33.4125 m
Clearly collision will take place between surfaces at D1+0.30m = 33.7125m