Question

3. (15 points) Assume you are agile enough to run across a horizontal surface at 8.50 m/s, independently of the value of the gravitational field. Assume your mass to be 90 kg. a) (5 points) What would be the radius of an airless spherical asteroid of uniform density 1.1 x 10 kg/m' on which you could launch yourself into orbit by running? b) (3 points) What would be the mass of this asteroid? c) (3 points) What would be your orbital period around the asteroid? d) (4 points) Using conservation of angular momentum, show that your running would not produce significant rotation of the asteroid if it was initially stationary. (lasteroid =MR)

Answer 1

a)

v = speed of person = 8.50 m/s

m = mass of the person = 90 kg

R = radius of the asteroid

V = Volume of asteroid = 4$\pi$R³/3

$\rho$ = density = 1100 kgm^-3

Mass of asteroid is given as

M = $\rho$V = 4$\pi$$\rho$R³/3

The gravitational force between the person and asteroid provides the necessary centripetal force, hence

Gravitational force = Centripetal force

GMm/R² = m v²/R

GM = v² R

4$\pi$$\rho$GR³/3 = v² R

4$\pi$$\rho$GR²/3 = v²

4 (3.14) (1100) (6.67 x 10^-11) R²/3 = 8.5²

R = 5266.55 m

b)

Mass of asteroid is given as

M = $\rho$V = 4$\pi$$\rho$R³/3

M = 4 (3.14) (1100) (5266.55)³/3

M = 6.7 x 10¹⁴ kg

c)

Using Kepler's third law

T² = 4$\pi$² R³/(GM)

T² = 4(3.14)² (5266.55)³/((6.67 x 10^-11)(6.7 x 10¹⁴))

T = 11354 sec

d)

w = angular speed of asteroid

w = m v R/((0.4) M R² )

w = (90) (8.5) (5266.55)/((0.4) (6.7 x 10¹⁴) (5266.55)² )

w = 5.42 x 10^-16​​​​​​​ rad/s