Answer to Question #201983 in Astronomy | Astrophysics for Manas Awasthi

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

(a) The maximum height corresponds to the zero velocity, so we may write the law of conservation of energy. We may assume the potential energy at the surface of the Earth is 0,

so the total mechanical energy is $E_1 = E_{k,1} = \dfrac{mv^2}{2}$ . At the maximum height $E_2 = E_{p,2} = mgh$ .

Due to the conservation of energy law, $E_1=E_2,$ so $\dfrac{mv^2}{2} = mgh,$

$h = \dfrac{v^2}{2g} = \dfrac{(2.8\cdot10^3)^2}{2\cdot9.81} = 4\cdot10^5\,\mathrm{m}.$

(b) More correct solution should take the change of g into account.

The potential energy at the distance r from the center of the Earth is $U(r) = -\dfrac{GM_{\oplus}m}{r}$ . Therefore, the law of conservation of energy takes form

$\dfrac{mv^2}{2} -\dfrac{GM_{\oplus}m}{R_{\oplus}} = 0 - \dfrac{GM_{\oplus}m}{R_{\oplus}+h}$ .

Therefore, $\dfrac{v^2}{2} = GM_{\oplus}\left(\dfrac{1}{R_{\oplus}}-\dfrac{1}{R_{\oplus}+h} \right)$ , $\;\;h = \dfrac{v^2R_{\oplus}^2}{2GM_{\oplus}-v^2R_{\oplus}}\\ h = \dfrac{(2.8\cdot10^3)^2\cdot(6371\cdot10^3)^2}{2\cdot 6.67\cdot10^{-11}\cdot5.97\cdot10^{24}- (2.8\cdot10^3)^2\cdot6371\cdot10^3} = 4.26\cdot10^5\,\mathrm{m}$