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Physics Energy/Universal Gravitation Mix?
The magnitude of the attractive force of gravity between two bodies is F = GMm/r^2. G is a constant equal to 6.67×10−11 N·m^2/kg^2, M and m are the masses, and r is the distance between the centers of the two bodies. The gravitational force of a star of mass 1.14×1032 kg and radius 3.48×109 m is the sole force acting on a rocket of mass 1.25×106 kg. The rocket is stationary relative to the star at distance of 1.85×1011 m. Sadly, the rocket has exhausted its fuel, and it will be pulled to its doom inside the star. How fast will it be moving when it reaches the surface of the star?
I keep getting 1.52 x 10^7 and cannot figure out where I went wrong. Your help is appreciated. Thanks. :-)
1 Answer
- MadhukarLv 71 decade agoFavorite Answer
The potential energy of the rocket on the surface of the planet
= - GMm/R, where M = mass of the planet, m = mass of the rocket and R = its radius.
Potential energy of the rocket at the distance h = 1.85x10^11 m
= - GMm/(R+h)
When the rocket reaches the surface of the planet, it loses potential energy
= -GMm/(R+h) - [- GM/R]
= GMm [1/R - 1/(R+h)]
= GMmh/[R(R+h)]
This potential energy of the rocket gets converted into its kinetic energy when it reaches the surface of the planet as its initial kinetic energy was zero at the height h.
=> (1/2)mv^2 = GMm/[R(R+h)]
=> v^2
= 2GM/[R(R+h)]
= 2 * (6.67x10^-11) * (1.14x10^32)/[(3.48x10^9)(3.48x10^9 + 1.85x10^11)]
= 15.2076 x 10^21/(6.559104 x 10^11)
=> v = 1.523 x 10^5 m/s.
