Question

The asteroid belt circles the sun between the orbits of mars and jupiter. one asteroid has a period of 5.0 earth years. assume a 365.25-days year and msun = 1.99 × 1030 kg. what is the asteroid's orbital radius?

Answer 1

Using Kepler's 3rd law which is: T² = 4π²r³ / GM 
Solved for r : 
r = [GMT² / 4π²]⅓ 
Where G is the universal gravitational constant,M is the mass of the sun,T is the asteroid's period in seconds, andr is the radius of the orbit.
Change 5.00 years to seconds : 
5.00years = 5.00years(365days/year)(24.0hours/day)(6... = 1.58 x 10^8s 
The radius of the orbit then is computed:
r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.58 x 10^8s)² / 4π²]⅓ = 4.38 x 10^11m 

Answer 2

The asteroid's orbital radius is about 4.37 × 10¹¹ m

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Further explanation

Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:

$\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }$

F = Gravitational Force ( Newton )

G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )

m = Object's Mass ( kg )

R = Distance Between Objects ( m )

Let us now tackle the problem !

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Given:

orbital period of asteroid = T = 5 × 365.25 × 24 × 3600 ≈ 1.58 × 10⁸ s

mass of Sun = M = 1.99 × 10³⁰ kg

Asked:

orbital radius = R = ?

Solution:

$\Sigma F = ma$

$G \frac{ M m} { R^2 } = m \omega^2 R$

$G \frac{ M } { R^2 } = \omega^2 R$

$G \frac{ M } { R^3 } = \omega^2$

$G \frac{ M } { R^3 } = (2\pi \div T)^2$

$R = \sqrt[3] { GM (\frac{T}{2\pi})^2 }$

$R = \sqrt[3] { 6.67 \times 10^{-11} \times 1.99 \times 10^{30} \times (\frac{1.58 \times 10^8}{2\pi})^2 }$

$R \approx 4.37 \times 10^{11} \texttt{ m}$

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Learn more

• Impacts of Gravity : https://brainly.com/question/5330244
• Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454
• The Acceleration Due To Gravity : https://brainly.com/question/4189441

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Answer details

Grade: High School

Subject: Physics

Chapter: Gravitational Fields