Question

What is the value of escape velocity of earth? Give the answer with steps pleeasee.... ☺​

Answer 1

SOLUTION.

$</u></strong></p><p><strong><u>let \: the \: mass \: of \: the \: object \: be \: m \\ mass \: of \: earth \: be \: M \\ \\ let \: we \: through \: the \: body \: with \: velocity \: v \\ at \: height \: h \\ its \: velocity \: is \: u \\ potential \: energy \: at \: height \: h = \frac{</u></strong></p><p><strong><u>[tex]let \: the \: mass \: of \: the \: object \: be \: m \\ mass \: of \: earth \: be \: M \\ \\ let \: we \: through \: the \: body \: with \: velocity \: v \\ at \: height \: h \\ its \: velocity \: is \: u \\ potential \: energy \: at \: height \: h = \frac{GMm}{h} \\ \\ acc \: to \: law \: of \: conservation \: of \: energy \\ \\ \frac{1}{2} m {v}^{2} = \frac{1}{2} m {u}^{2} + \frac{GMm}{h} \\ u \: is \: non \: zero \: hence \: \frac{1}{2} m {u}^{2} \: is \: a \: positive \: \\ term \: so \: if \: we \: remove \: it \: than \: inequality \: changes \\ \\ \frac{1}{2} m {v}^{2} \geqslant \frac{GMm}{h} \\ {v}^{2} \geqslant \frac{2GM}{h} \\ v \geqslant \sqrt{\frac{2GM}{h} } \\ where \: h = radius \: of \: earth \: for \: escape \: velocity \\ \\ put \: all \: values \: \\ \\ we \: get \: v \geqslant 11.2km {hr}^{ - 1}$

ESCAPE VELOCITY IS THE MINNIMUM VELOCITY REQUIRED TO MAKE A BODY OF ANY MASS LEAVE THE ATMOSPHERE OR EFFECT OF GRAVITATIONAL FIELD OF THE EARTH IF WE THROUGH A BODY WITH THIS VELOCITY THAN IT WILL NEVER BACK TO EARTH SURFACE.

Answer 2

Explanation:

On the surface of the earth the escape velocity is about 11.2 kmr which approx 33 times of speed of the sound

hope this will help u dear

or before answer will help u more than this