Suppose that the mirror described in Part A is initially at rest a distance R away from the sun. What is the critical value of area density for the mirror at which the radiation pressure exactly cancels out the gravitational attraction from the sun?

hey , the answer is c

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow

thermal energy is the answer

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft