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28 August, 21:46

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?

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Answers (1)
  1. 28 August, 22:32
    0
    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
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