Question

Scientists are developing a new space cannon to shoot objects from the surface of the Earth directly 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.
Note: Neglect the air resistance for this problem. Use 6.67×10−11 m3kg−1
s
−2
for the gravitational
constant, 6371 km for the Earth’s radius, and 5.97 × 1024 kg for the Earth’s mass.

Answer 1

Given:

u = 2.8 km/s

g = 9.81 m/$s^2$

G = 6.67×$10^{-11}m^3/kg$

M = 6371 km

r = 5.97 × $10^{24$ kg

To find:

1) h =?

2) H =?

Step-by-step explanation:

1)

The potential energy & kinetic energy near the Earth's surface is constant.

$\displaystyle E_{kinetic} = E _{poential}$

$\displaystyle \frac{1}{2}mv^2 = mgh$

where

m = mass of object

v = velocity

g = acceleration due to gravity

h = height

by deleting m

h = $\displaystyle \frac{v^2}{2g}$

  = $\displaystyle \frac{(2.8 \times 10^3)^2}{2 \times 9.81}$

h = 399.59 km

∴ The height of the projectile is 399.59 km.

2)

Given that, gravitation is not constant.

So the total work done is equal to the initial kinetic energy.

$\displaystyle \frac{1}{2}mv^2 = \int\limits{F} \, dr$

          = $\displaystyle \int\limits {mg'} \, dr$

          = $\displaystyle \int\limits^H_R {m\frac{GM}{r^2} } \, dr$

By solving the above equation,

H = 6797.92 km

∴ The required height is 6797.92 km.