Question

A projectile is fired vertically from Earth's surface with an initial speed of 4.0 km/s. Neglecting air drag, how far above the surface of Earth will it go?

Answer 1

Given Data

Initial speed, v = 4 km/s = 4 x 10^3 m/s

We know that

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

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

Gravitational Constant , G = 6.67 x 10^-11 m³ kg^-1 s^-2

To FInd :-

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

Multiply above eqn with 2

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

- 2* 6.67 x 10^-11 * 6 x 10^24 / ( R + h ) =    ( - 2 * 6.67 x 10^-11 * 6 x 10^24 / 6.4 x 10^6 ) + ( 4000 )^2

- 8 x 10^14 / ( R + h ) = - 1.25 x 10^8 + 1.6 x 10^7

- 8 x 10^14 / ( R + h ) = - 1.09 x 10^8

                          R + h = 7.338 x 10^6

                                 h = 7.338 x 10^6 - 6.4 x 10^6

                                   = 938911.17 m = 9.389 x 10^5 m

Max height reached from surface, h = 9.389 x 10^5 m