1.You are the science officer on a visit to a distant solar system. Prior to lan

Question

1.You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 18,000 km and its rotation period to be 22.3 hours. You have previously determined that the planet orbits 2.2 X 108 km from its star with a period of 402 earth days. Once on the surface you find that the free-fall acceleration is 12.2 m/s2.

What is the mass of the planet?

What is the mass of the star?

2.In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 2.7 X 1016 kg and a radius of 5.8 km.

What is the speed of a satellite orbiting 3.4 km above the surface?

What is the escape speed from the asteroid?

Explanation / Answer

1.solution

( mass of the planet

a = GM / R²
12.2 m/s² = 6.674*1011N·m²/kg² * M / (½*18*106m)² (convert km to m)   

M = 1.48*1025 kg

mass of the star

For "orbit", centripetal acceleration = gravitational acceleration, or
²r = (2/T)²r = 4²r / T² = v²/r = GM/r²
where the first four terms are all expressions for centripetal acceleration
and G = Newton's gravitational constant = 6.674*1011 N·m²/kg²
and M = mass of central body
and r = orbit radius
and T = orbit period = 402days * 86400s/day = 3.47*107 s

4²r / T² = GM/r² rearranges to
M = 4²r³ / GT² = 4² * (2.2*1011m)³ / (6.674*1011N·m²/kg² * (3.47*107s)²)
M = 5.22*1030 kg

2.solution

Given in SI units:

M = 2.7 X 1016 kg
G = 6.67428*10-11 m^3/kg-s^2 (Universal Gravitational Constant)
Ra = 5800 m
Ro = 3400 m

Rt = (5800 m) + (3400 m) = 9200 m

Orbital velocity above surface

v = SQRT { [GM] / Rt }
v = SQRT { [ (6.67428*10-11 m^3/kg-s^2) * (2.7 X 1016  kg) ] / [ 9200 m ] }
v = SQRT [ 195.86 m/s ]
v = 14 m/s

Escape velocity from the orbit of the satellite around the of the asteroid:

Ve = SQRT { [2GM] / Rt }
Ve = SQRT { [ 2 *(6.67428E-11 m^3/kg-s^2) * (2.7 X 1016  kg) ] / [ 9200 m ] }
Ve = SQRT { 391.72 m^2/s^2 }
Ve = 19.79 m/s

Escape velocity from the surface of the asteroid:

Ve = SQRT { [2GM] / Rt }
Ve = SQRT { [ 2 *(6.67428E-11 m^3/kg-s^2) * (2.7 X 1016  kg) ] / [ 5800 m ] }
Ve = SQRT { 621 m^2/s^2 }
Ve = 24.92 m/s

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