Question

Suppose that on earth you can jump straight up a distance of $45cm.$Asteroids are made of material with mass density $2800kg/m^3.$What is the maximum diameter of a spherical asteroid from which you could escape by jumping?

Short answer

Diameter,$D=4802meters$

Step-by-step solution

Given information

On earth you can jump straight up a distance of$36cm,h=36cm=0.36m$

Asteroids are made of material with mass density, $d=2800kg/m^3$

height of the jumprole="math" localid="1649935054690" $h=45cm=0.45m$

Explanation

Escape velocity is given by :

$$\begin{gathered} v=\sqrt{\frac{2GM}{R}} \\ R=\frac{2GM}{v^2} \end{gathered}$$

Where

$G$is the universal gravitational constant

$M$is the mass

Density, role="math" localid="1649934989575" $$\begin{gathered} d=\frac{M}{V} \\ d=\frac{M}{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3} \\ M=\frac{4}{3}d{\mathrm{\pi R}}^3=\left(\frac{4}{3}\right)\times \mathrm{\pi }\times 2800\times {\mathrm{R}}^3=11729{\mathrm{R}}^3 \end{gathered}$$

Now using conservation of energy as :

$$\begin{gathered} \frac{1}{2}m{v}^2=mgh \\ v=\sqrt{2gh}=\sqrt{2\times 9.8\times 0.45}=3.0m/s \end{gathered}$$

Plugging the values in the radius equation

$$\begin{gathered} R=\frac{2\times 6.67\times {10}^{-11}\times 11729R^3}{3^2} \\ R=\sqrt{\frac{9}{1.56\times {10}^{-6}}} \\ R=2402m \end{gathered}$$

Diameter of a spherical asteroid is role="math" localid="1649935494858" $D=2\times 2402=4804m$

So, the maximum diameter of a spherical asteroid from which you could escape by jumping is $4804m$Hence, this is the required solution.