Question

A projectile is fired vertically from Earth's surface with

aninitial speed of 10 km/s. Neglecting air drag, how far abovethe
surface of Earth will it go?

Answer 1

`h=25.52*10^6 m`

Step-by-step 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* G M / ( R + h ) = ( - 2 G M / R ) + v^2`

`-2*6.67*10^(-11)*6*10^(24)/ ( R + h )`

=
`( (- 2* 6.67*10^(-11)*6*10^(24)) /(6.4*10^6)} +10000^2`

=
`-2.50625*10^7 J`

`=- 8*10^(14) / ( R + h )=-2.50625* 10^7`

`R+h=31.92*10^(6)`

`h=31.92*10^(6)-6.4*10^6`

`h=25.52*10^6 m`