Question

The Earth slows in rotation by 1.4×103 seconds over around 100 years. If this energy is transferred to the Moon, how much further away does the Moon every year?

Answer 1

present time period of the Earth, T1 = 24 hours

= 24*60*60 s

= 86400 s

angular speed, w1 = 2*pi/T1

= 2*pi/(86400) rad/s

time period of the earth after 100 years, T2 = 24*60*60 + 1400 s

= 87800 s

we know, Me = 5.97*10^24 kg
Re = 6.370*10^6 m

The loss of rotataional kinetic energy, delta_KE = (1/2)*I*(w1^2 - w2^2)

= (1/2)*(2/5)*Me*Re^2*(w1^2 - w2^2)

= (1/5)*5.97*10^24*(6.37*10^6)^2*( (2*pi/86400)^2 - (2*pi/87800)^2)

= 8.106*10^27 J

present distance between the Earth and the moon, r = 384400 km

= 384400*10^3 m

let d is the increase in distance after 100 years.

we know, mass of the moon, M_moon = 7.346*10^22 kg

use, loss of rotational kinetic energy of the earth = gain in mechanical energy of the moon

delta_KE = -(1/2)*G*Me*M_mmon/(r + d) - ( -(1/2)*G*Me*M_mmon/r )

= (1/2)*G*Me*M_mmon/r - (1/2)*G*Me*M_mmon/(r + d)

= (1/2)*G*Me*M_moon*(1/r - 1/(r +d) )

8.106*10^27 = (1/2)*6.67*10^-11*5.97*10^24*7.346*10^22*(1/(384400*10^3) - 1/(384400*10^3 + d) )

==> d = 1.04*10^8 m <<<<<-----------------Answer

Note : here d represents increase in distance. If we need new distance, r_new = r + d