Question

Jump to JupiterThe gravitational potential energy of a 1kg object is plotted versus position from Earth’s surface to the surface of Jupiter. Mostly it is due to the two planets.

Make the crude approximation that this is a rectangular barrier of widthm and approximate height of$\mathbf{4}\mathit{X}{\mathbf{10}}^{\mathbf{8}}\mathbf{}\mathbf{j}\mathbf{/}\mathbf{kg}$. Your mass is 65 kg, and you launch your-self from Earth at an impressive 4 m/s. What is the probability that you can jump to Jupiter?

Short answer

The probability that one can jump to Jupiter is$e^{-2x{10}^{52}}$ using the concept of transmission probability.

Step-by-step solution

Tunneling  

The transmission or tunneling probability can be determined using transmitted intensity and the incident intensity.

In the case of tunneling barriers being wide, it can be found as follows.

$\mathit{T}\mathbf{\cong }\mathbf{16}\frac{\mathbf{E}}{{\mathbf{U}}_{\mathbf{0}}}\left(1-\frac{E}{U_0}\right){\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{L}\sqrt{\mathbf{2}\mathbf{m}\left(U_0-E_1\right)\mathbf{/}\mathbf{ħ}}}$

Here E is the jump energy, U₀is barrier energy, L is the length of the tunnel, and m is the mass of the particle.

Given quantities

The given values are: -

• length of the barrier,$L=6X{10}^{11}m$.
• Height of the barrier,$h=4X{10}^8\mathrm{J}/\mathrm{kg}$.
• The mass of the jumper is$m=65\mathrm{kg}$.
• Launch velocity is given as $v=4m/s$.

Finding the jump energy and the barrier energy

Estimate the jump energy for the launch velocity, v=4 m/sas:

$$\begin{gathered} E=\frac{1}{2}m{v}^2 \\ =\frac{1}{2}x65x4 \\ =520\mathrm{J} \end{gathered}$$

Find the barrier energy by multiplying the mass with potential height as:

$$\begin{gathered} U_0=65x4x{10}^8 \\ =2.6x{10}^{10}\mathrm{J} \end{gathered}$$

Finding the transmission probability for the jump

Estimate the value of transmission probability for the jump by putting the obtained values as:

$$\begin{gathered} T_1\cong 16\frac{E_1}{U_0}\left(\frac{E_1}{U_0}\right)e^{-2L\sqrt{2m\left(U_0-E_1\right)lħ}} \\ =16x\frac{520}{2.6x{10}^{10}}x\left(1-\frac{520}{2.6x{10}^{10}}\right)e^{-2x6x{10}^{11}\sqrt{2x65\left(2.{610}^{10}-520\right)/1.05x{10}^{-34}}} \\ \cong e^{-2x{10}^{52}} \end{gathered}$$

Therefore, the probability of the jump to Jupiter will practically be $e^{-2x{10}^{52}}$.