Answer to Question #202220 in Mechanics | Relativity for Aditya

Question

Scientists are developing a new space cannon to shoot objects from the surface of the Earth di-

rectly into a low orbit around the Earth. For testing purposes, a projectile is fired with an initial

velocity of 2.8 km/s vertically into the sky.

Calculate the height that the projectile reaches, ...

(a) assuming a constant gravitational deceleration of 9.81 m/s2

.

(b) considering the change of the gravitational force with height.

Answer 1

a)

$u=2.8km/s=2800m/s$

$g=9.81m/s^2$

$v^2=u^2+2as$

$v=0$

$0=2800^2-2\times9.81\times h$

$h=399.59km$

b)

here g is not constant

by energy conservation

$U=(\frac{2gh}{1+g/R})^{1/2}$

$u^2+\frac{u^2h}{R}=2gh$

$h(2g-u^2/R)=u^2$

$h=\frac{u^2R}{2gR-u^2}$

$h=\frac{2800^2 \times 6371\times 10^3}{2\times 9.81\times 6371\times 10^3-2800^2}=\frac{4.99\times 10^{13}}{124.87\times 10^6-7.84\times 10^6}=426.4km$