Physics 321 Quiz #5-Friday, Feb. 23 Consider a spherical asteroid made of rock w
ID: 1875342 • Letter: P
Question
Physics 321 Quiz #5-Friday, Feb. 23 Consider a spherical asteroid made of rock with density pa 2 g/cm3 and radius a. It is in a circular orbit of radius r about Saturn. A pebble of mass m lies on the surface of the asteroid, either at a point facing, or opposite the direction of Saturn. 1. In terms of G, M, r, a and 8m, express the tidal force on the pebble. The tidal force is the difference between the gravitational force from Saturn acting on the pebble vs the force it would experience if it were located at the center of the asteroid. Keep only the first order term when expanding in terms of a. Note that the tidal force is trying to lift the pebble from the surface. 2. In terms of G, a, om and pa, what is the gravitational force acting on the pebble due to the gravita- tional interaction with the asteroid. 3. Equate the two forces to find the radius r at which the tidal force rips apart the asteroid. Saturn's mass is M, 5.68 × 1026 kg. Note that G = 6.67 x 10-11 Nm2/kg2. Compare your answer to the radius of Saturn,R,-5.82 × 104 km. 4. What would happen if the asteroid were made of ice instead of rock? 5. Express your answer for c in terms of R,, Pa and the density of Saturn pa. Note the density of Saturn is 0.9 gm/cm3. It would float in your bathtub, if your bathtub were large enough. The average density of Jupiter is 1.3 g/cm3 and the average density of Earth is 5.5 g/cm3 6. If a small moon and planet slowly spiral toward each other, and if the moon and planet have the same density, will the moon be torn apart by tidal forces before reaching the surface of the planet?Explanation / Answer
form the given data
rhoa = 2 g/cm^3 = 0.002 kg / (0.01)^3 = 2*10^-3 * 10^6 = 2*10^3 kg/m^3
radius = a
orbit of radius r about the saturn
pebble mass = dm
1. tidal force = Fsurface - Fcenter
tidal force = GMs*dm(1/(r - a)^2 - 1/r^2)
Ft = GMs*dm((r - a)^-2 - r^-2)
Ft = GMs*dm(r^(-2)(1 + a/2r) - r^(-2))
Ft = GMs*dm*a/2r^3
2. gravitational force acting on pebble due asteroid = Fg
Fg = G(rhoa)(4*pi*a^3/3)*dm/a^2 = 4*pi*G(rhoa)dm*a/3
3. Ft = Fg
GMs*dm*a/2r^3 = G*rho*4*pi*a*dm/3
Ms/2r^3 = rho*4*pi/3
r^3 = 3Ms/8*rho*pi
Ms = 5.68*10^26 kg
r = 3.23643*10^7 m
4. if the asteroid were made of ice, then its density will be about 1000 kg/m^3 instead of 2000
and hence it will break apart at r' = r(2^1/3) = 4.07764966*10^7 m