Question

A projectile of mass 64 is shot from the surface of Planet Z with mass 4.34 x 1024 kg and radius 5.91 x 106 m by means of a very powerful cannon. If the projectile reaches a height of 3.08 x 105 m above Planet Z's surface, what was the speed of the projectile when it left the cannon?

Answer 1

`v = 2.2 * 10^3 m/s`

Step-by-step explanation:

Gravity on the surface of the planet is given as

`g = (GM)/(R^2)`

here we know that

`M = 4.34 * 10^(24) kg`

`R = 5.91 * 10^6 m`

now we have

`g = ((6.67 * 10^(-11))(4.34 * 10^(24)))/((5.91 * 10^6)^2)`

`g = 8.3 m/s^2`

now by energy conservation we have

initial KE + initial gravitational PE = final gravitational PE

`(1)/(2)mv^2 - (GMm)/(R) = - (GMm)/(R+h)`

`v^2 = 2((GM)/(R) - (GM)/(R + h))`

`v^2 = 2(6.67 * 10^(-11))(4.34 * 10^(24))((1)/(5.91 * 10^6) - (1)/(5.91 * 10^6 + 3.08 * 10^5))`

`v = 2.2 * 10^3 m/s`