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A planet orbits the Sun at a distance of 2.87×109 km. If the mass of the Sun is 1.99×1030 kg, find the orbital period of the planet.

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Answer:

2.65×10⁹ seconds, or 84.1 Earth years

Step-by-step explanation:

Don't forget to convert km to m before doing calculations.

Acceleration due to gravity = centripetal acceleration

GM / r² = v² / r

v² = GM / r

v² = (6.67×10⁻¹¹ m³/kg/s²) (1.99×10³⁰ kg) / (2.87×10¹² m)

v² = 4.62×10⁷ m²/s²

v = 6.80×10³ m/s

Period = circumference / velocity

T = 2πr / v

T = 2π (2.87×10¹² m) / (6.80×10³ m/s)

T = 2.65×10⁹ s

Converting to other units:

T = 7.37×10⁵ hr

T = 3.07×10⁴ days

T = 84.1 years

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