Question

A less than successful inventor wants to launch a s mall satellite into the orbit by launching them up from the surface of the earth at a very high speed. With what speed shall he launch the satellite if it has a speed of 500m/s at a height of 400km? Ignore air resistance.

Answer 1

One would be tempted to use an equation of motion here, but that wont take into consideration teh change in the acceleration of gravity due to increasing height. the way to go about this is through energy considerations, and that total energy is conserved. So equate the total energy (KE +PE) on the ground with that at a height of 400 km. Let the subscripts \(i\) and \(f\) denote initial and final velocities where the final velocity is at the height of 400 km. \[\frac{1}{2}mV_{i}^{2} -\frac{GM_{E}m}{R}=\frac{1}{2}mV_{f}^{2} -\frac{GM_{E}m}{R+r}\]where \(v\) is the velocity, \(G\) the gravitational constant, \(R\) the radius of the earth, and \(r\) the height of the rocket. Note that we must remember that the potential energy in this situation is negative, since it is defined as being equal to zero at infinity. Also, Note that the mass of the rocket \(m\) is not important as it cancels out leaving \[\frac{1}{2}V_{i}^{2} -\frac{GM_{E}}{R}=\frac{1}{2}V_{f}^{2} -\frac{GM_{E}}{R+r}\]. rearranging, we find that \[V_{i} =\sqrt{V_{f}^{2} +2GM_{E}\left(\frac{1}{R}-\frac{1}{R+r}\right)}\]

So just plug the numbers in, and you will have your answer.