sol11 - HW Solutions # 11- 8.01 MIT - Prof. Kowalski...

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Unformatted text preview: HW Solutions # 11- 8.01 MIT - Prof. Kowalski Universal Gravity . 1 ) 12.23 Escaping From Asteroid Please study example 12.5 "from the earth to the moon". a ) The escape velocity derived in the example (from energy conser- vation) is : v esc = 2 Gm A R A (1) Where: m A ≡ Asteroid’s mass=3.6 × 10 12 kg R A ≡ Asteroid’s radius=700 m G = 6 . 673 × 10 − 11 Nm 2 /kg 2 Plugging in these numbers into equation (1): v esc = 0 . 83 m/s You can certainly walk that fast on earth. However, you could not walk on the asteroid faster and faster to achieve this speed because you would go into orbit at a lower velocity ( v esc / √ 2) at which point the normal force of the ground would be zero so there would be no more friction to accelerate you! To get a feeling of how small this gravity is let’s do some estimation of the time it takes to reach this velocity on the asteroid. The gravitational force F g = Gm A m/R 2 A on the surface of planet for a mass, m ∼ 100 kg is ∼ 0.05 N. Let’s take µ = 1. The friction would be ∼ 0.05 N so the acceleration a ∼ . 05 / 100 = 0 . 0005 m/s 2 and the time t it takes to reach v esc = 0 . 83 m/s is: t ∼ . 83 . 0005 = 1660 s ∼ 30 min 1 b ) The question is about the comparison with vertical leap on earth. Using: v 2 y − v 2 y = − 2 g earth d we have: d = v 2 y 2 gd ≈ . 03 m d ≈ 3 cm Even octogenarians can jump ∼ 10 × this length. 2 2 ) 12.24 Satellite’s altitude and mass m S := Satellite’s mass. m E := Earth’s mass. R := The distance between the center of earth and satellite....
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This note was uploaded on 04/22/2011 for the course PHYS 1441 taught by Professor White during the Spring '08 term at UT Arlington.

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sol11 - HW Solutions # 11- 8.01 MIT - Prof. Kowalski...

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