Question

A huge cannon is assembled on an airless planet (ignore any effects due to the planet's rotation). The planet has a radius of 5.00 × 106 m and a mass of 3.95 × 1023 kg. The cannon fires a projectile straight up at 2000 m/s. An observation satellite orbits the planet at a height of 1000 km. What is the projectile's speed as it passes the satellite?

Answer 1

Answer:

The projectile's speed as it passes the satellite is 1497.8 m/s.

Explanation:

Given that,

Radius of planet $r=5.00\times10^{6}\ m$

Mass of planet $m=3.95\times10^{23}\ kg$

Speed = 2000 m/s

Height = 1000 km

We need to calculate the projectile's speed as it passes the satellite

Using conservation of energy

$E_{1}=E_{2}$

$\dfrac{1}{2}mv_{1}^2+\dfrac{GmM}{r_{1}}=\dfrac{1}{2}mv_{2}^2+\dfrac{GmM}{R+h}$

$\dfrac{v_{1}^2}{2}+\dfrac{GM}{r_{1}}=\dfrac{v_{2}^2}{2}+\dfrac{GM}{R+h}$

$-\dfrac{v_{2}^2}{2}=-(\dfrac{GM}{R}-\dfrac{GM}{R+h}-\dfrac{v_{1}^2}{2})$

$v_{2}^2=v_{1}^2+2GM(\dfrac{1}{R+h}-\dfrac{1}{R})$

$v_{2}=\sqrt{v_{1}^2+2GM(\dfrac{1}{R+h}-\dfrac{1}{R})}$

Put the value into the formula

$v_{2}=\sqrt{2000^2+2\times6.67\times10^{-11}\times3.95\times10^{23}(\dfrac{1}{5.00\times10^{6}+1000\times10^{3}}-\dfrac{1}{5.00\times10^{6}})}$

$v_{2}=1497.8\ m/s$

Hence, The projectile's speed as it passes the satellite is 1497.8 m/s.