Gravitational Time Dilation

Time passes more slowly by a factor of x x at plane cruising altitude of 12000 m 12000 \text{ m} above the earth's surface, compared to the time experienced by an object at infinity. At approximately how many places after the decimal point does x x differ from 1.000 ? 1.000\ldots?


Useful Constants:

  • The radius of the earth is 6.37 × 1 0 6 m 6.37 \times 10^6 \text{ m} .
  • The speed of light is 3 × 1 0 8 m / s 3 \times 10^8 \text{ m}/\text{s} .
  • Newton's gravitational constant is 6.67 × 1 0 11 N m 2 / kg 2 6.67 \times 10^{-11} \text{ N}\cdot \text{m}^2 / \text{kg}^2 .
  • The mass of the earth is 5.97 × 1 0 24 kg 5.97\times 10^{24} \text{ kg} .
1 4 10 30

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2 solutions

Matt DeCross
May 10, 2016

The factor that time runs slower compared to infinity is:

1 2 G M r c 2 . \sqrt{1-\frac{2GM}{rc^2}}.

Plugging in all of the numbers above, one finds 0.9999999993 0.9999999993 , which differs from 1 at the 10th place after the decimal point.

Peter Macgregor
Jul 24, 2016

I picked the right answer with a little bit of general knowledge. I know that

1. 1. Gravitational time dilation has been measured in the Earth's gravitational field.

2. 2. It is very difficult to measure, certainly much less than 1 part in ten thousand!

3. 3. The best atomic clocks have an accuracy many orders of magnitude poorer than 1 in 1 0 30 10^{30} .

Juggling these facts you can pick out the correct answer without knowing any formula or doing any calculations!

@Peter Macgregor I think that sort of reasoning is much more instructive than just bashing the formula! :)

Matt DeCross - 4 years, 10 months ago

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Thanks Matt.

I actually thought it was a bit cheeky!

Peter Macgregor - 4 years, 10 months ago

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