Question

A planet of mass 7.00 1025 kg is in a circular orbit of radius 6.00 1011 m around a star. The star exerts a force on the planet of constant magnitude 6.51 1022 N. The speed of the planet is 2.36 104 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?

Answer 1

Answer:

A) 1.88 * 10^17 m

B) 1.22 * 10^34 J

C) 1.95 * 10^34 J

Explanation:

Parameters given:

Mass of planet = 7.00 * 10^25 kg

Radius of orbit = 6.00 * 10^11 m

Force exerted on planet = 6.51 * 10^22 N

Velocity of planet = 2.36 * 10^4 m/s

A) The distance traveled by the planet is half of the circumference of the orbit (which is circular).

The circumference of the orbit is

C = 2 * pi * R

R = radius of orbit

C = 2 * 3.142 * 6.0 * 10¹¹

C = 3.77 * 10¹² m

Hence, distance traveled will be:

D = 0.5 * 3.77 * 10¹²

D = 1.88 * 10 ¹² m/s

B) Work done is given as:

W = F * D

W = 652 * 10²² * 1.88 * 10¹¹

W = 1.22 * 10³⁴ J

C) Change in Kinetic energy is given as:

K. E. = 0.5 * m * v²

K. E. = 0.5 * 7 * 10^25 * (2.36 * 10^4)²

K. E. = 1.95 * 10³⁴ J