On earth :
m = mass of jumper
h = height gained = 3 m
v = speed at the earth = ?
using conservation of energy
kinetic energy on earth = potential energy at the maximum height
(0.5) m v2 = mgh
v = sqrt(2gh)
v = sqrt(2 x 9.8 x 3)
v = 7.67 m/s
on the asteroid :
escape speed is given as
vs = sqrt(2GM/R) where M = mass of the asteroid , R = radius of the asteroid
V = Volume of asteroid = 4R3/3
= density = 2500
kg/m3
we know that : M = V
M = 4R3
/3
vs = sqrt(4R3
G/3R)
vs = sqrt(4R2
G/3)
7.67 = sqrt(4R2(2500)(6.67
x 10-11)/3)
R = 9177.4 m
d = diameter = 2R = 2 x 9177.4 = 18354.8 m
Asteroid jumping (problem 68/chapter 8) [6] As a member of the 2040 Olympic committee, you're considering...
2. Asteroid jumping (problem 68/chapter 8) 16] As a member of the 2040 Olympic committee, you're considering a new sport: asteroid jumping. On Earth, world-class high jumpers routinely clear 2m. Your job is to make sure athletes jumping from asteroids will return to the asteroid. Make the simplifying assumption that asteroids are spherical, with average density 2500kg/m. For safety, make sure even a jumper capable of 3m on Earth with return to the surface. What do you report for the...