## Related questions with answers

To escape Earth’s gravitational field, a rocket must be launched with an initial velocity called the escape velocity. A rocket launched from the surface of Earthhas velocity v (in miles per second) given by v = √2GM/r + v_0^2 - 2GM/R ≈ √192000/r + v_0^2 - 48 where v_0 is the initial velocity, r is the distance from the rocket to the center of Earth, G is the gravitational constant, M is the mass of Earth, and R is the radius of Earth (approximately 4000 miles). (c) A rocket launched from the surface of a planet has velocity v (in miles per second) given by v = √10600/r + v_0^2 - 6.99. Find the escape velocity for this planet. Is the mass of this planet larger or smaller than that of Earth? (Assume that the mean density of this planet is the same as that of Earth.)

Solution

VerifiedSetting $v\to0, r\to \infty$ we get $0+v_0^2-6.99\to 0\Rightarrow v_0=\sqrt{6.99}$.

For Earth we have $\dfrac{GM}{R}=24$, for the other planet we have $\dfrac{GM}{R}=3.495$.

Assuming that $\dfrac{G}{R}$ is the same constant for both planet then the mass the of Earth is bigger.

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