A rocket is launched straight up from the earth’s surface at a speed of 1.90×10^4 m/s?

A rocket is launched straight up from the earth’s surface at a speed of 1.90×10^4 m/s. What is its speed when it is very far away from the earth? Answer in m/s

Initial Kinetic Energy (KE) of the rocket = 1/2*mv^2 = 1.805m x 10^8 J
Initial potential energy (PE) = -GMm / R
= – (6.67 * 10^-11) (6 * 10^24) m / (6.4 * 10^6)
= – 6.25m * 10^7 J

Total initial energy = KE + PE = 1.18m x 10^8 J

From Conservation of Energy,
Total Initial Energy = Total Final Energy

So
Total Final Energy = 1.18m x 10^8

Final PE = 0 (Put r = infinity in the expression for PE)
So
Final KE = 1.18m * 10^8

1/2*m*v^2 = 1.18m * 10^8

v = 1.54 x 10^4 m/s

Hope this helps.

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What confuses me is that there are two variables, I understand r —> ∞ therefore the Gravitational Potential Energy —-> 0, but it seems you threw the mass variable in the equation and you were able to solve for velocity without mass even given! How did you do that?

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