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 81.0%. How many liters of water would be sufficient fuel to very slowly push the Moon 2.10 mm away from the Earth? The density of water is Pwater = 1.00kg/liter, the Earth's mass is Mearth = 5.97 x 10 kg, the Moon's mass is Mmoon = 7.36 x 1022 kg, and the separation of the Earth and Moon is dem = 3.84 x 10 m. water: Liters

Answer 1

The gravitational potential energy between tow Spherical object with center to center distance being is given by çm, miz U- where m & m2 of respective obsects. are the masses mron for earth system, when distance between them is the potential energy dem ç m Earth Mmoon U = dem at a the moon is placed distance (demtd) from the Center of Earth, then the Potential energy G MEarth Mmoon U2 = (de,mtd)

needed so, Work done w = U₂ -U camearth Maroon) * Wa w = (a mearth Maroon Em td. dem » W= GM earth Mamoon 1 dem (demtd) we know that, G=6.67710" N.m? N.m? ugn? give Mearah = 5.97x1024 kg veni 22 M moon = 7.36*10² kg dem = 3.84 x 108m 8 d = 2.1mm = 2.616?m * wa (6.67*15", 5.97*1024 27.3661022) » [cesarity 24. 1. v)-ię savy J 4

W = 4.173380367 x 107 J ņ have if we of water v liters Water then mass of this much -pu soj input energy, E=mc² E = pve? so, the efficiency W = w pvce n = 3114 E pc2r ginen, p=1kg/liter n 817.5 we have, w = 4.17338036771017, we know that e=3x108 81 Too mis & 1017 liters V ) = B 4.173380367 810 1 81 (3x108) 8 loo V = 5,724801601 litery ? V 5.725 liters (appron) >