NOTE: This problem has a small tolerance instead of the usual 2.0%. Enter your a
ID: 2214838 • Letter: N
Question
NOTE: This problem has a small tolerance instead of the usual 2.0%. Enter your answers to four significant figures. Ito's Planet is in a solar system having a very few heavy elements; as a result, Ito's Planet has almost no elements heavier than oxygen. As a consequence, though Ito's Planet is about as large as Earth (the radius of Ito's Planet is 6.6780 106 m), the mass of Ito's Planet is much less than the mass of Earth (the mass of Ito's Planet is only 1.5434 1024 kg). Ito's planet also spins about its axis more rapidly than does Earth, making one complete spin every 10.0 hours. Although Ito's Planet has no magnetic field, those two regions at which the spin axis intersects with the surface of the Planet are still referred to as the polar regions, with one region designated as the North Pole and the other as the South Pole; Ito lives in the shelter of a crater at the South Pole. In a science fiction story, Ito is being visited by a young Captain Kirk. For this problem, use G = 6.6743 10-11 N?m2/kg2 for the universal gravitation constant. (a) During a visit to a research station on the equator of Ito's Planet, the Captain attempts to measure his Ito's Planet weight by stepping on a spring balance (such as a bathroom scale). The reading on the spring scale is 136.92 N. What is the Captain's mass? HINT: You must take into account the fact that Ito's Planet is spinning to correctly calculate the Captain's mass. kg (b) When the Captain returns to Ito's house at the South Pole and steps on a spring scale, he finds that the reading is larger than his measured weight at the equator. What is the Captain's measured weight at the South Pole of Ito's Planet? N.Explanation / Answer
M = 1.5434*10^24 kg
R = 6.6780*10^6 m
G = 6.6743*10^-11
on ito's planet,g = GM/R^2 = 2.3099 m/s^2
w = 2/[10*3600] = 1.7453*10^-4 rad/s
centrifugal acceleration,ac = w^2*R = 0.2034 m/s^2
downward acceleration on captain at equator,g' = g-ac = 2.1065 m/s^2
let mass of captain be m,mg' = 136.92 => m = 64.9988 kg Ans(a)
at north pole downward acceleration is g' = g
thus measured weight at pole,W = mg' = mg = 150.1407 N Ans(b)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.