Question

The asteroid belt circles the sun between the orbits of Mars and Jupiter. One asteroid has a period of $5.0$earth years. What are the asteroid’s orbital radius and speed?

Short answer

The asteroid’s orbital radius is$4.365\times {10}^{11}m$and its speed is$1.74\times {10}^{4}m/s.$

Step-by-step solution

Given information

The asteroid belt circles the sun between the orbits of Mars and Jupiter. One asteroid has a period of $5$earth years.

Calculation

The radius can be calculated using the time period and it is given by :

$$\begin{gathered} T=\frac{2\pi R}{v} \\ R=\frac{Tv}{2\pi } \\ R=\frac{(1.5768s\times {10}^{8}s)v}{2\pi } \\ R=(2.5\times {10}^7)v\_\_\_\_\_\_\left(1\right) \end{gathered}$$

Sun's gravitational attraction provides the asteroid's centripetal force and it is given by :

$$\begin{gathered} \frac{G{M}_s{M}_a}{R^2}=\frac{M_a\times v^2}{R} \\ {v}^2=\frac{G\times M_s}{R}\_\_\_\_\_\_\left(2\right) \end{gathered}$$

Continuation of calculation

Substituting the value of R in equation (2) we get,

$$\begin{gathered} v^2=\frac{G\times M_s}{2.5\times {10}^{7}v} \\ {v}^3=\frac{G\times M_s}{2.5\times {10}^7} \\ {v}^3=\frac{(6.67\times {10}^{-11}N{m}^2/k{g}^2)(1.989\times {10}^{30}kg)}{2.5\times {10}^7} \\ {v}^3=5.287\times {10}^{12}{m}^3/s^3 \\ v=1.74\times {10}^4\mathrm{m}/\mathrm{s} \end{gathered}$$

Substituting the value of 'v' in equation (1) we get,

$$\begin{gathered} R=(2.5\times {10}^{7}s)(1.74\times {10}^{4}m/s) \\ R=4.365\times {10}^{11}\mathrm{m} \end{gathered}$$

Final answer

The asteroid’s orbital radius is $1.74\times {10}^4\mathrm{m}/\mathrm{s}$and its speed is$4.365\times {10}^{11}\mathrm{m}.$