Chapter 13: Q13-93P (page 385)

Planet Roton, with a mass of $7.0 \times 10^{24} k g$ and a radius of $1600 k m$ , gravitationally attracts a meteorite that is initially at rest relative to the planet, at a distance great enough to take as infinite.The meteorite falls toward the planet. Assuming the planet is airless, find the speed of the meteorite when it reaches the planet’s surface.

Expert verified

The speed of the meteorite when it reaches the planet's surface $V = 2.4 \times 10^{4} \frac{m}{s}$

Step by step solution

The mass of planet Roton is $M = 7.0 \times 10^{24} k g$ .

The radius of planet Roton is

$\begin{array}{rcl} R & = & 1600 k m \\ = & 1.6 \times 10^{6} m \end{array}$

Using the law of conservation of energy, we can find the speed of the meteorite when it reaches the planet's surface.

Formulae:

$K_{1} + U_{1} = K_{2} + U_{2} K = \frac{1}{2} m v^{2} U = - \frac{G M m}{r}$

We have the equation for conservation of energy as

$K_{1} + U_{1} = K_{2} + U_{2}$

Now, the meteor starts from rest, which means $v_{1} = 0 o r K_{1} = 0$

Also, initially the meteor is at an infinite distance, so $U_{1} = 0$

So, the conservation of energy equation becomes

$K_{2} + U_{2} = 0 \frac{1}{2} m v_{2}^{2} - \frac{G M m}{r_{2}} = 0 v_{2} = \sqrt{\frac{2 G M}{R}}$

Hence

$V = 2.4 \times 10^{4} \frac{m}{s}$

The speed of the meteorite when it reaches the planet's surface, $V = 2.4 \times 10^{4} \frac{m}{s}$ .

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