A planet with mass 5.37×1024kg and radius 6.51×106m orbits its sun, a distance 141.68×109m away, in a uniform circular motion w/ velocity 3.3×104m/s. Assuming the planet is a uniform solid sphere spinning on its axis and has the same length of day as Earth, what is the orbital angular momentum of this planet (in kg⋅m2/s)?
To find the orbital angular momentum of the planet, we can use the formula:
L = mvr
where L is the angular momentum, m is the mass of the planet, v is the velocity, and r is the radius of the orbit.
Given:
Mass of the planet (m) = 5.37×10^24 kg
Radius of the orbit (r) = 141.68×10^9 m
Velocity (v) = 3.3×10^4 m/s
Now we can substitute these values into the formula:
L = (5.37×10^24 kg) × (3.3×10^4 m/s) × (141.68×10^9 m)
Simplifying:
L = (5.37 × 3.3 × 141.68) × (10^24 × 10^4 × 10^9) kg⋅m^2/s
L = 2344.0832 × (10^24 × 10^4 × 10^9) kg⋅m^2/s
L = 2344.0832 × (10^24 + 4 + 9) kg⋅m^2/s
L = 2344.0832 × 10^37 kg⋅m^2/s
Therefore, the orbital angular momentum of the planet is approximately 2.34 × 10^40 kg⋅m^2/s.