Question

what would the radius (in mm) of the earth have to be in order for the escape speed of the earth to equal the speed of light (300000000 m/s)? you may ignore all other gravitational interactions for the rocket and assume that the earth-rocket system is isolated.

Answer 1

The escape speed from a spherical object can be calculated as v_esc = √(2GM/R), where G is the gravitational constant, M is the mass of the object, and R is the radius.

Setting v_esc = c, the speed of light, and solving for R:

R = GM/c^2 = (6.67 x 10^-11 m^3 kg^-1 s^-2)(5.97 x 10^24 kg)/(3 x 10^8 m/s)^2

R = 1.20 x 10^8 m = 1,200,000,000 mm

So the radius of the Earth would need to be approximately 1,200,000,000 millimeters (or 1.2 x 10^8 meters) for its escape speed to equal the speed of light.