Question

Derive expression for escape speed​

Answer 1

Answer:

From this equation, it can be said that the escape velocity depends on the radius of the planet and the mass of the planet only and not on the mass of the body. Escape Velocity of Earth= 11.2 km/s. This was the derivation of the escape velocity of earth or any other planet.

Answer 2

Hi friend______,

• Before answering the question, I am going to give some of the basic Introduction.

For Example:

★ Considering you are a Strong Boy/Girl. If you are throwing a ball without energy, it will fall nearby you. And next, you have throwing the next ball after drinking some juice. It will fall some certain distance than the 1st ball. By seeing this you have become angry and you are throwing the 3rd ball after eating healthy foods.

This happens due to the presence of gravitslational force present in the earth.

★ From this we can conclude Newtons 3rd Law: "For every action, there is an equal and opposite reaction", which means that- there will be a equal number of action based on the amount of energy.

Answer:

Similarly, if we want the satellite to revolve around the earth, we want to launch the rocket in some particular speed. While launching the rocket, there are two forces acting on it. They are kinetic energy and potential energy.

Derivation:

$\rm E_{i}= K.E+P.E$

We know that,

$\rm Kinetic \: Energy = \frac{1}{2}mv^2_{i}$

so,

$\rm \: = \frac{1}{2}mv^2_{i}-\frac{GMMe}{Re}$

★ This is for Initial stage of rocket.

But,

At final stage:

$\rm E_{f}=0$

According to Law of Conservation:

$\rm E_{i} = E_{f}$

$\rm \frac{1}{2}mv^2_{i}-\frac{GMMe}{Re}=0$

$\rm \frac{1}{2}mv^2_{i}= \frac{GMMe}{Re}$

$\rm mv^2_{i}=\frac{2GMMe}{Re}$

$\boxed{\rm v^2_{e}= \frac{2GMe}{Re}}=> \rm Escape\: Velocity$

We know that:

$\boxed{ \rm \: g=\frac{GMe}{Re}}$

We can further convert it as:

$\boxed{\rm Re^2= GMe}$

Then:

$\rm = \frac{2Re^2g M_{E}}{M_{E}R_{E}}$

$\rm V_{e}^{2}=2Reg$

$\rm V_{e} = √2Reg$

= 11.2 km/sec