Suppose two planets (spherical in shape) in radii R and 2R, but mass M and 9M respectively have a centre to centre separation 8 R as shown in the figure. A satellite of mass 'm' is projected from the surface of the planet of mass 'M' directly towards the centre of the second planet. The minimum speed 'v' required for the satellite to reach the surface of the second planet is √(a/7GM/R) then the value of 'a' is ____________.
[Given : The two planets are fixed in their position]

Solution

Physics sir · 4 steps

Past the neutral point the 9M planet pulls it in, so the satellite just needs to reach that point.

1. Find the neutral point. Let the two pulls balance at distance x from the centre of M. Then GM/x² = 9GM/(8R - x)², so 8R - x = 3x and x = 2R. This point lies in the gap between the two surfaces.
2. Why this point. Before x = 2R the net pull is back towards M. After it, the net pull is towards 9M. A satellite that just reaches the neutral point with zero speed then falls onto the second planet, so that is the minimum-speed case.
3. Potentials. At the surface of M (R from M, 7R from 9M): V₁ = -GM/R - 9GM/7R = -16/7GM/R. At the neutral point (2R from M, 6R from 9M): V₂ = -GM/2R - 3GM/2R = -2GM/R.

   1/2mv² = m(V₂ - V₁) = m(-2 + 16/7)GM/R = 2/7mGM/R

4. So. v² = 4/7GM/R, so v = √(4/7GM/R) and a = 4.

The common slip: 2. Forgetting the factor 1/2 in kinetic energy, and writing v² equal to the potential difference, gives a = 2, not 4.

The idea to keep. Between two fixed masses, a body only has to be pushed as far as the neutral point. Equate the launch kinetic energy to the rise in potential energy up to that point.

Answer from the NTA final answer key. Solution by PrepWiser.