This problem is closely related to the spectacular impact of the comet Shoemaker-Levy with Jupiter in July 1994:

http://www.jpl.nasa.gov/sl9/ sl9.html

A rock far outside our solar system is initially moving very slowly relative to the Sun, in the plane of Jupiter’s orbit around the Sun. The rock falls towards the Sun, but on its way to the Sun it collides with Jupiter. Calculate the rock’s speed just before colliding with Jupiter. Explain your calculation and any approximations that you make.

$M_{sun}=2\times {10}^{30}kg$,$M_{Juipter}=2\times {10}^{27}kg$

Distance, Sun to Jupiter $=8\times {10}^{11}m$

Radius of Jupiter$1.4\times {10}^{8}m$

The given data can be listed below as,

• The mass of the Sun is,$M_{sun=}2\times {10}^{30}kg$
• The mass of the Jupiter is,$M_{juipter}=2\times {10}^{27}kg$

  
  

• The distance of the Sun to the Jupiter is,$d=8\times {10}^{11}m$
• The radius of the Jupiter is,$1.4\times {10}^{8}m$

The creation and destruction of any kind of energy is not possible but it can only be transformed from one kind to another. In other words, it can be said that the total amount of energy contained in a system remains constant. It can be expressed as follows,

PE + KE…(1)

Here, PE is the potential energy of the system and KE is the kinetic energy of the system.

The expression for the gravitational potential energy is as follows,

$U=-\frac{GmM}{R}$…(2)

Here, G is the gravitational constant that is$6.67\times {10}^{-11}N{m}^2/k{g}^2$, m is the mass of first object, M is the mass of second object and r is distance between first and second object.

The expression for the kinetic energy is as follows

$KE=\frac{1}{2}m{v}^2$…(3)

Here, m is the mass of the object and v is the velocity with which the object is moving.

Write the expression for the energy conservation of the given system using equation (1).

$-\frac{Gm{M}_{Sun}}{d}-\frac{Gm{M}_{juipter}}{r}+\frac{1}{2}m{v}^2=0$

Here, m is the mass of the rock and v is the speed of rock before a collision with the planet.

Rearrange the above expression.

$$\begin{gathered} -\frac{Gm{M}_{Sun}}{d}-\frac{Gm{M}_{juipter}}{r}+\frac{1}{2}m{v}^2=0 \\ G\left(\frac{M_{Sun}}{d}+\frac{M_{Juipter}}{r}\right)=\frac{1}{2}{v}^2 \\ v=\sqrt{2G\left(\frac{M_{Sun}}{d}+\frac{M_{Juipter}}{r}\right)} \end{gathered}$$

Substitute all the values in the above expression.

$$\begin{gathered} v=\sqrt{2\left(6.67\times {10}^{11}N.m^2/k{g}^2\right)\times \frac{1kg.m/s^2}{1N}\left(\frac{2\times {10}^{30}kg}{8\times {10}^{11}m}+\frac{2\times {10}^{29}kg}{1.4\times {10}^{8}m}\right)} \\ =4.73\times {10}^{4}m/s \end{gathered}$$

Thus, the speed of rock before collision with Jupiter is $4.73\times {10}^{4}m/s$.