Question

Derive an expression for escape speed.​

Answer 1

Answer:

The unit of break speed or getaway speed is communicated in meter each seconds (m.s-1). This is additionally the SI unit of getaway speed.

Explanation:

The determination of getaway speed is characterized as far as an item and its speed. At the point when the article moves with a speed at which the number-crunching complete of the item's motor energy, its gravitational potential energy compares to nothing. This implies the item ought to have more prominent dynamic energy than the gravitational possible energy to run away to vastness.

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,

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

★ 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:

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

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

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

= 11.2 km/sec