Formula used:
KE=GMm2r
For energy in radius
ri=∣∣∣−GMm2r∣∣∣=nk
where, n is integer and k is constant. Now, according to the given situation,
For r=R
GMm2R=nk …..(i)
For energy in radius
rf=∣∣∣−GMm2r∣∣∣=(n−1)k
According to the given situation,
For r=3R2
GMm2(3R2)=(n−1)k …..(ii)
Solving (i) and (ii), we get
k=GMm6R
Now this implies that for Rmax
GMm2Rmax=(1)GMm6R
Rmax=3R
Final Answer: 3