Question

Consider a spherical asteroid with a radius of 20 km and a mass of 9.25 ✕ 1015 kg. Assume the asteroid is roughly spherical. (a) What is the acceleration due to gravity on the surface of this asteroid? m/s2 (b) Suppose the asteroid spins about an axis through its center, like the Earth, with a rotational period T. What is the smallest value T can have before loose rocks on the asteroid's equator begin to fly off the surface? h

Answer 1

a)
Use equation;
g = GM / R²
Where g is acceleration , G is universal gravitational constant, M is mass and R is radius....
g = (6.67 * 10^-11)(8.70*10^15 kg) / (20000 m)²
g = 580290 / 400000000
g = 0.00144285034.........OR
g = 1.444216 * 10^-3 m/s²
................................
b)
As it is a sphere so the circumference of this sphere would be;
C = 2πR
As Time = Distance / Velocity
T = 2πR / V
We can find V by;
g = V² /R
V² = gR
=> V = √gR
Now, the equation becomes;
T = 2πR/ √g/R
T = 2πR * √R/g
T = 2(3.1415926)(20000) * √ 20000/ 0.0014242585
T = 131947 * √15959266
T = 125663.81** 3910
T = 491345497 seconds = 491345497 / 3600 hours = hours
===> T = 15.7966 years