Question

Space explorers discover an $8.7\times {10}^{17}kg$asteroid that happens to have a positive charge of $4400C$. They would like to place their $3.3\times {10}^{5}kg$spaceship in orbit around the asteroid. Interestingly, the solar wind has given their spaceship a charge of $–1.2C$. What speed must their spaceship have to achieve a $7500-km$-diameter circular orbit?

Short answer

Their spaceship has to achieve a speed of$7.33m/s$.

Step-by-step solution

Given Information

Charge of asteroid$=q_1=4400C$

Mass of asteroid$=m_1=8.7\times {10}^{17}kg$

Charge of their spaceship$=q_2=-1.2C$

Mass of their spaceship$=m_2=3.3\times {10}^{5}kg$

Orbital radiusrole="math" localid="1648453557483" $=r=7500/2=3750km=3750\times {10}^{3}m$

Calculation

The electrostatic force of attraction between these two interacting bodies can be given from the following equation:

$F_e=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{\left|q_1\right|\left|q_2\right|}{r^2}$

Where,

$\frac{1}{4{\mathrm{\pi \epsilon }}_0}=8.99\times {10}^{9}N{m}^2/C^2$

By substituting the given values in the above equation, we get,

$$\begin{gathered} F_e=\frac{8.99\times {10}^9\times 4400\times 1.2}{{\left(3750\times {10}^3\right)}^2} \\ {F}_e=3.37N \end{gathered}$$

The gravitational force of attraction between these two interacting bodies can be given from the following equation:

$F_g=G\times \frac{m_1{m}_2}{r^2}$

By substituting the given values in the above equation, we get,

$$\begin{gathered} F_g=6.67\times {10}^{-11}\times \frac{8.7\times {10}^{17}\times 3.3\times {10}^5}{{\left(3750\times {10}^3\right)}^2} \\ {F}_g=1.36N \end{gathered}$$

Hence, the total force of attraction can be given as:

$$\begin{gathered} F=F_e+F_g \\ F=3.37+1.36 \\ F=4.73N \end{gathered}$$

Now, this attractive force will balance the centrifugal force.

We know that the centrifugal force is given as:

$F=\frac{m{v}^2}{r}$

Hence,

$4.73=\frac{m{v}^2}{r}$

By rearranging the terms to get velocity,

$$\begin{gathered} v=\sqrt{\frac{4.73\times r}{m}} \\ v=\sqrt{\frac{4.73\times 3750\times {10}^3}{3.3\times {10}^5}} \\ v=7.33m/s \end{gathered}$$

Final answer

Hence, the speed required for their spaceship to achieve a $7500-km$-diameter is $7.33m/s$.