1) You are the science officer on a visit to a distant solar system. Prior to la
ID: 1560775 • Letter: 1
Question
1) You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.8 × 107 m and its rotation period to be 22.3 hours. You have previously determined that the planet orbits 7.4 × 1011 m from its star with a period of 402 earth days. Once on the surface you find that the acceleration due to gravity is 26.4m/s2. What are the mass of (a) the planet and (b) the star? (G = 6.67 × 10-11 N · m2/kg2)
2) Spaceman Speff orbits spherical asteroid X with his spaceship. To remain in a circular orbit at 356 km from the asteroid's center, he should maintain a speed of 20 m/s.What is the mass of asteroid X? (G = 6.67 × 10-11 N · m2/kg2)
Spaceman Speff orbits spherical asteroid X with his spaceship. To remain in a circular orbit at km from the asteroid's center, he should maintain a speed of m/s.What is the mass of asteroid X? (G = 6.67 × 10-11 N · m2/kg2)
1)(a) 3.2 × 1025 kg, (b) 2.0 × 1032 kg 2)(a) 3.2 × 1025 kg, (b) 1.2 × 1032 kg 3)(a) 5.8 × 1025 kg, (b) 2.0 × 1032 kg 4)(a) 5.8 × 1025 kg, (b) 1.2 × 1032 kgExplanation / Answer
1)
Hence, 1)(a) 3.2 × 1025 kg, (b) 2.0 × 1032 kg is the correct ans.
explanation
g = GM/R^2
M = g R^2/G
R = D/2 = 1.8 x 10^7/2 ; g = 26.4 m/s^2
M = 26.4 x (9 x 10^6)^2/6.67 x 10^-11 = 3.21 x 10^25 kg
Hence, M = 3.21 x 10^25 kg
For star, Fg = Fc
G M m /r^2 = m v^2/R
M = v^2 R/G
v = 2 pi R/t = 2 x 3.14 x 7.4 x 10^11/402 x 86400 = 1.34 x 10^5 m/s
M = (1.34 x 10^5)^2 x 7.4 x 10^11/6.67 x 10^-11 = 2 x 10^32
Hence, M = 2 x 10^32
Hence, 1)(a) 3.2 × 1025 kg, (b) 2.0 × 1032 kg is the correct ans.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.