Question

Show that the escape velocity of a body from the earth's surface is root 2 times its velocity in circular orbit just above the earth's surface.

Answer 1

Answer:

To prove:

Escape Velocity is √2 times of orbital Velocity.

Proof:

Orbital velocity of satellites is given as :

$orbital \: v = \sqrt{ \frac{Gm}{r} }$

Where m => mass of Earth , r => radius of Earth.

Escape Velocity can be calculated by the us of Conservation of Energy theorem:

KE1 + PE1 = KE2 + PE2

$\frac{1}{2} M{(v. \: esc)}^{2} - \frac{GmM}{ {r}^{2} } = 0 + 0$

$= > v \: esc. = \sqrt{ \frac{2Gm}{r} }$

$= > v \: esc. = \sqrt{2 } \times ( \sqrt{ \frac{Gm}{r} } )$

$= > v \: esc. = \sqrt{2} \times (orbital \: v)$

[Hence proved.]

Answer 2

To prove:

Escape Velocity is √2 times of orbital Velocity.

Proof:

Orbital velocity of satellites is given as :

$\bold\green{orbital \: v = \sqrt{ \frac{Gm}{r}}}$

m = mass of Earth

r = radius of Earth

Escape Velocity can be calculated by the us of Conservation of Energy theorem:

KE1 + PE1 = KE2 + PE2

$\tt{\frac{1}{2} M{(v. \: esc)}^{2} - \frac{GmM}{ {r}^{2} } = 0 + 0}$

$\tt{v \: esc. = \sqrt{ \frac{2Gm}{r}}$

$tt{ v \: esc. = \sqrt{2 } \times \sqrt{ \frac{Gm}{r}}$

$v \: esc. = \sqrt{2} \times orbital \: vel. }$