Question

Find the orbital speed of an ice cube in the rings of Saturn. The mass of Saturn is 5.68 x 1026 kg, and use an orbital radius of 3.00 x 105 km. (G = 6.67 × 10-11 N ∙ m2/kg2)

Answer 1

The orbital speed of an ice cube in the rings of saturn is approximately 3.56 * 10^6 m/s

What is law of gravitation?

The law of gravitation states that the force of gravitation is directly proportional to the product of the masses and inversely proportional to the distance between the masses. Mathematically;

F = GMm/r²

where

M and m are the mass of ice cube and

Recall that;

s = Gm1/r^2

Also;

F = sm²

Substitute to have;

s = m²/F

For the centripetal acceleration

a = v²/r

Such that;

v²/r = Gm/r²

v² = Gm/r

v = √Gm/r

Substitute the given parameters into the formula to have:

V =  √6.67×10^-11 *  5.68 x 10^26 / 3.00 x 10^5

V = 355358.97m/s = 3.56 * 10^6 m/s

Therefore the orbital speed of an ice cube in the rings of saturn is approximately 3.56 * 10^6 m/s

Learn more on orbital speed here: brainly.com/question/7260440

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