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.

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

Explanations & Calculations

a)

• Since g is constant, consider the mechanical energy conservation of the canon between near-earth surface & the maximum height.

$\qquad\qquad \begin{aligned} \small E_{kinetic}&=\small E_{potential}\\ \small \frac{1}{2}mv^2 &=\small mgH\\ \small H&=\small \frac{v^2}{2g}\\ &=\small \frac{(2.8\times10^{3})^2}{2\times9.81}\\ &=\small 399.59\,km \end{aligned}$

b)

• As long as the gravitation is not constant, total work can be taken into consideration.
• Total work done as it goes up equals the initial kinetic energy

$\qquad\qquad \begin{aligned} \small \frac{1}{2}mv^2 &=\small \int F dr =\int mg' dr\\ &=\small \int_{R_e}^H m\frac{GM}{r^2 }.dr\\ &=\small GmM \int\frac{1}{r^2} dr\\ &=\small GmM \bigg[-\frac{1}{r}\bigg]_{R_e}^H\\ \small H&=\small 6797.32 \, km \end{aligned}$