- #31
ensign_nemo
- 29
- 9
Just as a sanity check, I took the figures given in your reference for the total energy dissipated by tidal forces:
"P = 110 ± 25 GW. For comparison this is 22 times smaller than the M% dissipation rate in the oceans (Cartwright & Ray 1991), but an order of magnitude greater than the rate in the atmosphere (Platzman 1991)."
https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20000072440.pdf
I multiplied 110 GW by 23, to combine losses from solid and liquid friction, which gives an estimate of the total amount of energy dissipated by all tidal forces on Earth:
http://www.wolframalpha.com/input/?i=110+*+10^9+*23+W
This gives 2.53 TW, about the same as total electrical power consumption on Earth.
I then used 1/2 * (G * Mass_earth * Mass_sun / Orbital radius of Earth):
http://www.wolframalpha.com/input/?...sun)*(mass+of+Earth))/(2*(astronomical+unit))
This gives 2.649 x 10^33 Joules for the orbital kinetic energy of Earth. Wolfram helpfully adds that this figure is 98% of their number for the "Earth's orbital kinetic energy around the Sun", so it looks right.
Divide the second number by the first number:
http://www.wolframalpha.com/input/?.../(2*(astronomical+unit))+/+(110+*+10^9+*23+W)
The result is that the Earth will stop orbiting the Sun due to its loss of orbital kinetic energy from tidal friction in a mere 33 trillion years.
If we rather cavalierly assume constant tidal friction over time, the age of the Earth is 4.5 billion years, so we have lost about
http://www.wolframalpha.com/input/?i=(age+of+the+earth)+/+33+trillion+years
or 1.38 * 10 ^ -4 = 0.000138 of the initial kinetic energy.
Call it about one percent of one percent.
That's also tossing all lunar tidal friction into the equation, which is not accurate for our solar distance calculation, as that mostly slows down the moon, rather than the Earth.
So, yeah ... it's not a big factor.
"P = 110 ± 25 GW. For comparison this is 22 times smaller than the M% dissipation rate in the oceans (Cartwright & Ray 1991), but an order of magnitude greater than the rate in the atmosphere (Platzman 1991)."
https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20000072440.pdf
I multiplied 110 GW by 23, to combine losses from solid and liquid friction, which gives an estimate of the total amount of energy dissipated by all tidal forces on Earth:
http://www.wolframalpha.com/input/?i=110+*+10^9+*23+W
This gives 2.53 TW, about the same as total electrical power consumption on Earth.
I then used 1/2 * (G * Mass_earth * Mass_sun / Orbital radius of Earth):
http://www.wolframalpha.com/input/?...sun)*(mass+of+Earth))/(2*(astronomical+unit))
This gives 2.649 x 10^33 Joules for the orbital kinetic energy of Earth. Wolfram helpfully adds that this figure is 98% of their number for the "Earth's orbital kinetic energy around the Sun", so it looks right.
Divide the second number by the first number:
http://www.wolframalpha.com/input/?.../(2*(astronomical+unit))+/+(110+*+10^9+*23+W)
The result is that the Earth will stop orbiting the Sun due to its loss of orbital kinetic energy from tidal friction in a mere 33 trillion years.
If we rather cavalierly assume constant tidal friction over time, the age of the Earth is 4.5 billion years, so we have lost about
http://www.wolframalpha.com/input/?i=(age+of+the+earth)+/+33+trillion+years
or 1.38 * 10 ^ -4 = 0.000138 of the initial kinetic energy.
Call it about one percent of one percent.
That's also tossing all lunar tidal friction into the equation, which is not accurate for our solar distance calculation, as that mostly slows down the moon, rather than the Earth.
So, yeah ... it's not a big factor.