Gregory Emery

2022-01-07

Between Mars' and Jupiter's orbits, the asteroid belt revolves around the sun. A single asteroid orbits the earth every 5.0 years.
What are the orbital radii and velocities of the asteroids?

Jimmy Macias

Expert

Given values:
${T}_{1}=5years=5\cdot 365\cdot 24\cdot 3600s=1.5768\cdot {10}^{8}s$
$G=6.67\cdot {10}^{-11}\frac{Nk{g}^{2}}{k{g}^{2}}$
${m}_{sun}=1.99\cdot {10}^{30}kg$
The fravitational force between the sun and the orbiting asteroid is:
$F=G\frac{{m}_{sun}{m}_{a}}{{r}_{1}^{2}}$
The asteroid is affected by centripetal acceleration because it moves in a circular orbit
$F=m{a}_{c}$,
${a}_{c}=\frac{{v}^{2}}{r}$
According to the Newtons

trisanualb6

Expert

Using Kepler's 3rd law which is: ${T}^{2}=\frac{4{\pi }^{2}{r}^{3}}{GM}$
Solved for r:
$r={\left[\frac{GM{T}^{2}}{4{\pi }^{2}}\right]}^{\frac{1}{3}}$
Where G is the universal gravitational constantM is the mass of the sun, T is the asteroid's period in seconds, and r is the radius of the orbit.
Change 5.00 years to seconds :
5.00 years=5.00 years (365days/year)(24.0hours/day)($6\dots =1.58×{10}^{8}s$
The radius of the orbit then is computed:
$r={\left[\frac{\left(6.67×{10}^{-11}N\cdot \frac{m}{k{g}^{2}}\right)\left(1.99×{10}^{30}kg\right){\left(1.58×{10}^{8}s\right)}^{2}}{4{\pi }^{2}}\right]}^{\frac{1}{3}}=4.38×{10}^{11}m$

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