Question

Planet Roton, with a mass of $\mathbf{7}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{24}}\mathbf{}\mathit{k}\mathit{g}$and a radius of $\mathbf{1600}\mathbf{}\mathit{k}\mathit{m}$ , 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.

Short answer

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

Listing the given quantities

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

The radius of planet Roton is

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

Understanding the conceptof law of conservation of energy

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

Formulae:

$$\begin{gathered} K_1+U_1=K_2+U_2 \\ K=\frac{1}{2}m{v}^2 \\ U=-\frac{GMm}{r} \end{gathered}$$

Step 3: Calculations for speed of meteorite when it reaches the planet surface

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 $v_1=0or{K}_1=0$

Also, initially the meteor is at an infinite distance, so$U_1=0$

So, the conservation of energy equation becomes

$$\begin{gathered} K_2+U_2=0 \\ \frac{1}{2}m{v}_2^2-\frac{GMm}{r_2}=0 \\ {v}_2=\sqrt{\frac{2GM}{R}} \end{gathered}$$

Hence

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

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