Get the solution to "Between Mars' and Jupiter's orbits, the asteroid belt..."

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Gregory Emery

Answered question

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?

Answer & Explanation

Jimmy Macias

Beginner2022-01-08Added 30 answers

Given values:

T_{1} = 5 y e a r s = 5 \cdot 365 \cdot 24 \cdot 3600 s = 1.5768 \cdot 10^{8} s

G = 6.67 \cdot 10^{- 11} \frac{N k g^{2}}{k g^{2}}

m_{s u n} = 1.99 \cdot 10^{30} k g

The fravitational force between the sun and the orbiting asteroid is:

F = G \frac{m_{s u n} 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

Beginner2022-01-09Added 32 answers

Using Kepler's 3rd law which is:

T^{2} = \frac{4 π^{2} r^{3}}{G M}

Solved for r:

r = {[\frac{G M T^{2}}{4 π^{2}}]}^{\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 \times 10^{8} s

The radius of the orbit then is computed:

r = {[\frac{(6.67 \times 10^{- 11} N \cdot \frac{m}{k g^{2}}) (1.99 \times 10^{30} k g) {(1.58 \times 10^{8} s)}^{2}}{4 π^{2}}]}^{\frac{1}{3}} = 4.38 \times 10^{11} m

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