Question

A projectile is launched from the surface of a planet (mass = 15 x 1024 kg, radius = R = 9.6 x 106 m). What minimum launch speed is required if the projectile is to rise to a height of 6R above the surface of the planet? Disregard any dissipative effects of the atmosphere. Put your answer in km/s. ​

Answer 1

Answer:

The minimum launch speed is 13366.40 m/s.

Explanation:

Given that,

Mass of planet $M=15\times10^{24}\ kg$

Radius $R= 9.6\times10^{6}\ m$

Height = 6R

We need to calculate the speed

Using relation between gravitational force and total energy

$-\dfrac{GMm}{6R+R}=\dfrac{1}{2}mv^2-\dfrac{GMm}{R}$

$\dfrac{}{}mv^2=\dfrac{GMm}{7R}-\dfrac{GMm}{R}$

$\dfrac{1}{2}mv^2=\dfrac{6GMm}{7R}$

$v^2=\dfrac{2\times6GM}{7R}$

Put the value into the formula

$v=\sqrt{\dfrac{2\times6\times6.67\times10^{-11}\times15\times10^{24}}{7\times9.6\times10^{6}}}$

$v=13366.40\ m/s$

Hence, The minimum launch speed is 13366.40 m/s.

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