Question

how to derive an expression for escape velocity of satelite??

Answer 1

Use equations in Gravitation \[Escape Velocity=\frac{ GMm }{ r^{} }=\frac{ 1 }{ 2 }mv ^{2} \rightarrow v=\sqrt{\frac{ 2GM }{ r }}\]

The reason is because the potential energy, extremely far away from the center of the Earth is going to be extremely small. As the potential energy decreases, in normal situations, the kinetic energy should increase. But, in this case, the potential energy decreases because of the large distance and, therefore, the kinetic energy also decreases to zero. Since both are zero, you can equate them to find the escape velocity equation.

Answer 2

Sorry for rezzing an old post but I just have to comment. The previous poster was right, you do need to use energy to solve this problem however gravitational potential energy is negative. It is always negative no matter how far from a gravitational source you are. By moving further away from the earth gravitational energy becomes less negative and kinetic energy becomes less positive. By asking what is escape velocity what you are really asking is what velocity do I need to have in order to have zero net energy taking into account gravitational potential and kinetic energy.
\[U _{g}+K=0\]
\[-\frac{ GmM }{ d }+\frac{ 1 }{ 2 }mv^2=0\]
Yes, the math was right but it is an important distinction to make. This is why when you fall you gain velocity; you are losing gravitational potential energy.

Answer 3

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