Question

A projectile is fired vertically from Earth's surface with

Answer 1

Answer:

$h=25.52\times10^6 m$

Explanation:

Initial speed, v = 10 x 10^3 m/s

Mass of the earth, M = 6 x 10^24 kg

Radius of the earth, R = 6.4 x 10^6 m

Maximum from the surface of earth, h = ?

Let  m = Mass of the projectile

Solution:

Potential energy at maximum height =  ( Potential + Kinetic energy ) at the surface

$-G M m / ( R + h )=- G M m / R + (1/2) m v^2$

$- G M / ( R + h ) = - G M / R + (1/2) v^2$

$-2\times G M / ( R + h ) = ( - 2 G M / R ) + v^2$

$-2\times6.67\times10^{-11}\times6\times10^{24}/ ( R + h )$

=$( (- 2\times 6.67\times10^{-11}\times6\times10^{24}) /(6.4\times10^6)} +10000^2$

=$-2.50625\times10^7 J$

$=- 8\times10^{14} / ( R + h )=-2.50625\times 10^7$

$R+h=31.92\times10^{6}$

$h=31.92\times10^{6}-6.4\times10^6$

$h=25.52\times10^6 m$