Question

a 2.0 x 106 kg rocket is launched from the surface of the earth. what is the escape speed (in km/s) of the rocket with respect to its gravitational interaction with the sun? the initial distance of the rocket from the sun is 1.5 x 1011 m and the mass of the sun is 1.99 x 1030 kg. you may ignore all other gravitational interactions for the rocket and assume that the system is isolated.

Answer 1

The escape speed is calculated using the formula v_escape = sqrt(2GM/r), where G is the gravitational constant, M is the mass of the sun, and r is the initial distance of the rocket from the sun.

Plugging in the values:

v_escape = sqrt(2 * 6.67 x 10^-11 * (1.99 x 10^30 kg) / (1.5 x 10^11 m)) = sqrt(2 * 6.67 x 10^-11 * 1.99 x 10^30 kg / (1.5 x 10^11 m)) = sqrt(2 * 6.67 x 10^-11 * (1.99 / 1.5) x 10^30 kg / 10^11 m) = sqrt(2 * 6.67 x 10^-11 * 1.326 x 10^30 kg / 10^11 m) = sqrt(2 * 6.67 x 10^-11 * 1.326) x sqrt(10^30 kg / 10^11 m) = 2.8 x 10^4 m/s

So, the escape speed of the rocket with respect to its gravitational interaction with the sun is approximately 28,000 m/s or 28 km/s.