65.9k views
1 vote
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.

User Joar
by
8.0k points

1 Answer

5 votes

Answer:

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.

User SteveDeFacto
by
9.1k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.