## 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 Earth has velocity (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). (b) A rocket launched from the surface of the moon has velocity v (in miles per second) given by v = √1920/r + v_0^2 - 2.17. Find the escape velocity for the moon.

Solution

VerifiedAgain we set $r\to \infty, v\to 0$. Putting them in the equation we get $0+v_0^2-2.17\to 0\Rightarrow v_0=\sqrt{2.17}$

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