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

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.

1. What is the asteroid's orbital radius?

Express your answer in two significant figures.

What is the asteroid's orbital speed?

Express answer in two significant figures.

Solutions

Expert Solution

1 Earth year = 365.25 days
Ms = mass of the sun = 1.99 x 10^30 kg
T = the asteroid's orbital period = 5.0 years = 157,788,000 s
G = universal gravitational constant = 6.67 x 10^-11 N∙m²/kg²
r = the asteroid's orbital radius = to be determined

(a) What is the asteroid's orbital radius?

T = 2π√(r³/GMs)
T² = [2π√(r³/GMs)]²
T² = 4π²r³/GMs
T²GMs = 4π²r³
T²GMs/4π² = r³
r³ = T²GMs/4π²
r = ³√(T²GMs/4π²)

r = ³√[(157,788,000 s)²(6.67 x 10^-11 N∙m²/kg²)(1.99 x 10^30 kg)/4π²]
r = ³√[3.304660528415952 x 10^36/4π²]
r = 437,443,907,377 m ≈ 4.37 x 10^8 km ANSWER

(b) What is the asteroid's orbital speed?

v = orbital speed of the asteroid = to be determined

v = √(GMs/r)
v = √[(6.67 x 10^-11 N∙m²/kg²)(1.99 x 10^30 kg)/(437,443,907,377 m)]
v = √(303,428,617)
v = 17,419 m/s ≈ 17.4 km/s ANSWER

genius_generous answered 1 year ago

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