Suppose that you have found a way to convert the rest energy

Suppose that you have found a way to convert the rest energy of any type of matter directly to usable energy with an efficiency of 61.0%. How many liters of water would be sufficient fuel to very slowly push the Moon 3.10 mm away from the Earth? Use the following values for the density of water rho_water = 1.00 kg/liter, the Earth\'s mass M_earth = 5.97 times 10^24 kg, the Moon\'s mass M_moon = 7.36 times 10^22 kg, and the separation of the Earth and Moon d_E, M = 3.84 times 10^8 m.

Solution

Here ,

mass of earth , Mearth = 5.97 *10^24 Kg

mass of moon , Mmoon = 7.36 *10^22 Kg

d = 3.84 *10^8 m

energy used to push 3.1 mm , E = -G * Mearth * Mmoon * (1/(3.84 *10^8 + 0.0031) -(1/(3.84 *10^8))

E = -6.673 *10^-11 * 5.97 *10^24 * 7.36 *10^22 * (1/(3.84 *10^8 + 0.0031) -(1/(3.84 *10^8))

E = 1.54 *10^18 J

let the mass of the water is m

0.61 * m* c^2 = E

0.61 * m * (3 *10^8)^2 = 1.54 *10^18

m = 28.1 Kg

as density of water = 1 kg/litre

volume of water = 28.1 litre

the mass of water found is 28.1 litre