Question

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 65.0 % . How many liters of water would be sufficient fuel to very slowly push the Moon 2.50 mm away from the Earth? The density of water is ρ water = 1.00 kg/liter , the Earth's mass is M earth = 5.97 × 10 24 kg , the Moon's mass is M moon = 7.36 × 10 22 kg , and the separation of the Earth and Moon is d E,M = 3.84 × 10 8 m .

Answer 1

M_earth = mass of earth = 5.97 x 10²⁴ kg

M_moon = mass of moon = 7.36 x 10²² kg

d_E,M = initial separation between earth and moon = 3.84 x 10⁸ m

r = final separation between earth and moon = 3.84 x 10⁸ + 2.50 x 10^-3

Energy required to separate the earth and the moon is given as

E = U_f - U_i = (- G M_earth M_moon /r) - (- G M_earth M_moon /d_E,M )

E = G M_earth M_moon ​​​​​​​/d_E,M - G M_earth M_moon /r

= efficiency = 0.65

V = Volume of water

= density of water = 1 kg/L

mass of water is given as

M = V

M = 1 V

M = V

c = speed of light = 3 x 10⁸ m/s

rest energy of water is given as

E_rest = M c²

Given that

E = E_rest

G M_earth M_moon ​​​​​​​/d_E,M - G M_earth M_moon /r = M c²

(6.67 x 10^-11) (5.97 x 10²⁴)(7.36 x 10²²)​​​​​​​/(3.84 x 10⁸) - (6.67 x 10^-11) (5.97 x 10²⁴) (7.36 x 10²²)/(3.84 x 10⁸ + 2.50 x 10^-3) = (0.65) M (3 x 10⁸ )²

M = 8.5 L