Question

NEET PHYSICS CLASS 11
Plz solve it with solution. Question is in attachment...
answer is (45GM/4a)^1/2​

Answer 1

Answer:

All we need to do is give object kinetic energy exactly equal to difference in gravitional potential energy

Let the mass of object be m

$\frac{1}{2} m {v}^{2} \\ = \{ G \frac{16Mm}{2a} + G \frac{Mm}{8a} \} \\ - \: \{G \frac{16Mm}{9a} + G \frac{Mm}{a} \}\\ \\ \implies \\ {v}^{2} = 2G \frac{M}{a} (8 + \frac{1}{8} - \frac{16}{9} - 1) \\ \\ {v}^{2} = 2G \frac{Mm}{a} ( \frac{576 + 9 - 128 - 72)}{72} \\ \\ {v}^{2} = G \frac{Mm}{a} ( \frac{385}{36} ) \\ \\ v = \sqrt{G \frac{385M}{36a} }$

I think it is right answer

Answer 2

Explanation:

• Let P be the point on line joining the centres. The net field at that point is zero.
• Now GM/ r^2 – G16M / (10a – r)^2 = 0
• So (10a – r)^2 = 16 y^2
• Or 10a – r = 4r
• Or r = 2a
• Now potential at point P will be
• Vp = - GM/ r – G16M/ (10a – r)
•     = - GM (10 a – r) – r G16M / r(10 a – r)
•    = - GM(10 a – 2a) – 2a G16 M / 2a(10 a – 2a)
•    = - 5GM / 2a
• It will be able to reach the smaller planet if the particle projected from larger planet has adequate energy to cross the point.
• So the kinetic energy imparted to the body must be able to raise its total mechanical energy which is equal to potential energy at a point.
• So 1/2 mv^2 – G(16 M)m / 2a – GMm / 8a = mvp    
• Or v^2 / 2 – 8 GM / a – GM / 8a = 5 GMm / 2a
• Or v^2 = 45 GM / 4a
• Or v = √45 GM/ 4a                                                                                                                                     Or v = (45 GM / 4a )^1/2