Question

Neptune goes around the Sun in about 164.8 Earth years.

It maintains an orbit 4.5 x 1012 m away from the Sun. (MN = 1.024 x 1026 kg)
How much centripetal force keeps Neptune in its orbit?

Answer 1

The centripetal force that keeps Neptune in its orbit around the Sun is approximately 2.23 x 10^43 N.

Step-by-step explanation:

The centripetal force that keeps Neptune in its orbit around the Sun can be calculated using the formula F = (m * v^2) / r, where F is the centripetal force, m is the mass of Neptune, v is its orbital velocity, and r is the distance from Neptune to the Sun.

Given that the mass of Neptune (m) is 1.024 x 10^26 kg, and the distance from Neptune to the Sun (r) is 4.5 x 10^12 m, we can calculate the orbital velocity of Neptune using the formula v = 2 * π * r / T, where T is the orbital period of Neptune.

Substituting the values and solving the equation, we find that the orbital velocity of Neptune is approximately 5504 m/s. Plugging this value into the centripetal force formula, we can calculate that the centripetal force keeping Neptune in its orbit is approximately 2.23 x 10^43 N.