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Two planets have masses M and 16 M and their radii are a and 2a, respectively. The separation between the centres of the planets is 10a. A body of mass m is fired trom the surface of the larger planet towards the samller planet along the line joining their centres. For the body to be able to reach at the surface of smaller planet, the minimum firing speed needed is :

A. S
B. 4√GM/a
C. √GM²/ma
D. 3/2√5GM/a

User Axifive
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Final answer:

To find the minimum firing speed needed for a body to reach the surface of the smaller planet, use the principle of conservation of mechanical energy. The correct answer is C. √GM²/ma. Option C is correct.

Step-by-step explanation:

To find the minimum firing speed needed for a body to reach the surface of the smaller planet, we can use the principle of conservation of mechanical energy. The initial mechanical energy of the body consists of its kinetic energy and gravitational potential energy. The final mechanical energy is just the gravitational potential energy of the body on the smaller planet. Setting these two energies equal to each other, we can solve for the minimum firing speed.

First, we assume that there is no energy loss due to air resistance or other factors. This allows us to neglect factors like angle of launch and atmospheric drag.

Next, we can express the initial kinetic energy of the body as (1/2)mv^2, where m is the mass of the body and v is the firing speed.

The initial gravitational potential energy is given by -GM(16M)/10a, where G is the universal gravitational constant, M is the mass of the larger planet, and a is its radius. The negative sign accounts for the fact that the gravitational potential decreases as the body moves away from the surface of the planet.

The final gravitational potential energy on the smaller planet is given by -GM(16M)/2a.

Setting the initial and final energies equal to each other, we can solve for the minimum firing speed v.

The correct answer is: C. √GM²/ma.

User Themerius
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