(100%) Problem 1: A planet of mass 5.00 × 1025 kg is in a circular orbit of radi
ID: 3308803 • Letter: #
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(100%) Problem 1: A planet of mass 5.00 × 1025 kg is in a circular orbit of radius 2.00 x 1011 m around a star. The star exerts a force on the planet of constant magnitude 6.70 x 1023 N. The speed of the planet is 5.18 x 104 m/s 25% Part (a) In half a year the planet goes halfway around the star. What is the distance in meters that the planet travels along the semicirclc? Grade Summary Deductions Potential distancc 0% 100% tan acos) Submissions Attempts remaining: 4 (0% per attempt) detailed view sinO cosO cotanasin) atan acotansinh0 cosh() tanh cotanh( 0 DegreesRadians 0 CLEAR Submit I give up! Hints: 0 for a 0% deduction. Hints remaining: 0 Feedback: 0% deduction per feedback. 25% Part (b) During this half year", how much workin Joules is done on the planet by the gravitational force acting on the planet? là 2500 Part(c) What is the change in kinetic energy in Joules of the planet? 25% Part(d) What is the magnitude of the change of momenturm in kg·m/s of the planet?Explanation / Answer
(A) distance = pi r
= pi (2.00 x 10^11) m
= 6.28 x 10^11 m
(B) gravitational force is always perpendicular to the motion.
hence work done = 0
(C) Work done = change in KE = 0
(D) after half way, velocity vector will be in opposite direction that of intial direction.
vf = 5.18 x 10^4 m/s then vi = - 5.18 x 10^4 m/s
change in momentum = m (vf -vi)
= (5 x 10^25) (2 x 5.18 x 10^4)
= 5.18 x 10^30 kg m/s
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