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The aim is partly pedagogical. I want to assemble some examples illustrating how the google calculator can play a role in learning cosmology and related. Suppose you don't remember the Schw. radius of the earth.
Paste in "2G*mass of earth/c^2" press return, and you get 0.887 cm.
Gravitational time dilation is an important part of understanding cosmology. How much slower does a clock on Earth surface run than one "at infinity" i.e. effectively out of Earth's gravitational field? Call the rate of the distant clock "1" and subtract off the rate of the slower clock. The calculator knows the radius of earth. Paste in:
1 - (1 - .887 cm/radius of earth)^.5
You get about 0.7 parts per billion. Instead of typing "radius of earth" I'll assume we remember it is 6371 km.
Compared with one at sea level, how much faster does a clock run which is at an altitude of 1 km?
Paste in:
((1 - .887 cm/6372 km)/(1 - .887 cm/6371 km))^.5 - 1
You get about 1.1 x 10-13
A cesium clock runs about 9 billion cycles per second. After 1000 seconds the high clock is about one cycle ahead of the low clock.
What if the high clock is at an altitude of 100 meters? Paste in:
((1 - .887 cm/6371.1 km)/(1 - .887 cm/6371 km))^.5 -1
You get about 1.1 x 10-14
So after 10,000 second the high clock is about one cycle ahead.
That is how time varies in a gravitational field. The calculator is near the limit of its accuracy but since the field is nearly uniform in that range we can extrapolate and say that if the high clock is at an altitude of 10 meters it runs faster by about one part in 1.1 x 10-15.
So that after 100,000 seconds the high clock is about one cycle ahead. A day is 86,400 seconds so it takes a bit over a day to gain a cycle. But 10 meters is not much difference in altitude.
That is an example of how you use the ratio of the Schwarzschild radius of the earth, to its actual radius, to calculate something interesting namely a difference in the rates of passage of time.
Paste in "2G*mass of earth/c^2" press return, and you get 0.887 cm.
Gravitational time dilation is an important part of understanding cosmology. How much slower does a clock on Earth surface run than one "at infinity" i.e. effectively out of Earth's gravitational field? Call the rate of the distant clock "1" and subtract off the rate of the slower clock. The calculator knows the radius of earth. Paste in:
1 - (1 - .887 cm/radius of earth)^.5
You get about 0.7 parts per billion. Instead of typing "radius of earth" I'll assume we remember it is 6371 km.
Compared with one at sea level, how much faster does a clock run which is at an altitude of 1 km?
Paste in:
((1 - .887 cm/6372 km)/(1 - .887 cm/6371 km))^.5 - 1
You get about 1.1 x 10-13
A cesium clock runs about 9 billion cycles per second. After 1000 seconds the high clock is about one cycle ahead of the low clock.
What if the high clock is at an altitude of 100 meters? Paste in:
((1 - .887 cm/6371.1 km)/(1 - .887 cm/6371 km))^.5 -1
You get about 1.1 x 10-14
So after 10,000 second the high clock is about one cycle ahead.
That is how time varies in a gravitational field. The calculator is near the limit of its accuracy but since the field is nearly uniform in that range we can extrapolate and say that if the high clock is at an altitude of 10 meters it runs faster by about one part in 1.1 x 10-15.
So that after 100,000 seconds the high clock is about one cycle ahead. A day is 86,400 seconds so it takes a bit over a day to gain a cycle. But 10 meters is not much difference in altitude.
That is an example of how you use the ratio of the Schwarzschild radius of the earth, to its actual radius, to calculate something interesting namely a difference in the rates of passage of time.