A planet of mass 7.00e25 kg is in a circular orbit of radius 6.00e11 m around a

Question

A planet of mass 7.00e25 kg is in a circular orbit of radius 6.00e11 m around a star. The star exerts a force on the planet of constant magnitude 7.82e22 N. The speed of the planet is 2.59e4 m/s.

(a) In half a "year" the planet goes half way around the star. What is the distance that the planet travels along the semicircle?
distance =    m
(b) During this half "year", how much work is done on the planet by the gravitational force acting on the planet?
work =   J
(c) What is the change in kinetic energy of the planet?
?K =  J
(d) What is the magnitude of the change of momentum of the planet?

Explanation / Answer

a) perimeter of orbit = 2 * pi * radius
half year travel = half that = pi * radius

b) work = force * distance. Calculate force using your gravity equation (mass1, mass2, distance) and multiply by rsult in (a)

c) correct

d) This question is tricky- some would say nothing, becase the planet didn't increase or decrease speed. BUT momentum, by its definition, has a vector. The magnitude stayed the same, but the direction changed 180 degrees. So, you have to calculate the original momentum (mass * velocity, both of which you are given) and double that to find the change.

That's why the question is related to a 'half year'. The change of momentum in a year would be zero.

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