- #1
Labrodor
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The center of mass of the Earth is approximately at the center. The center of mass is also where the force vector of the gravitational force is pointing. (down that is).
Now the total gravitational force excerted on me should be the sum of the force excerted by the Earth plus anything else, especially the moon. With the sum of these two vectors being what I feel as "down", should the direction of down change as the moon's position in the sky changes?
I thought I'd do a quick calculation and hear if you guys think it makes sense.
Approximate average distance from the surface of the Earth to the center of the moon is
378 000 km. It has a mass of 7.34*10^22 kg.
The gravitational constant is G = (6.67*10^-11)*(m^3)*(kg^-1)*(s^(- 2))
Which should give me the acceleartion:
a_moon = G * Moon_mass / distance to moon^2
a_moon = (6.67*10^-11)*(m^3)*(kg^-1)*(s^(- 2)) * 7.34*10^22 kg / ((3.78*10^8 m)^2)
a_moon = 3.426x10^-5 m/s^2
With Earth's acceleration at 9.8 and the moon at the horizon we can do some quick trigonometry on the vectors. With the two vectors and their sum a_moon_vector + a_earth_vector forming a triangle, with the sum equal to the hypotenuse, we get an angle A which is the angle between the direction of felt down and the direction towards the centre of the earth.
Tan A = a_moon/a_earth
A = Tan^-1 ( 3.42*10^-5 / 9.8) = ca 2*10^-4 degrees
Is this enough to measured? Since we've got the ratio of the lengths in the triangle, we quickly find that if we have a pendulum hanging from a 300 meter long string (say, from the eiffel tower) it will deviate by 1 millimeter. That is, a change in 2 millimeters as the moon passes from one side of the horizon to the next.
Now the total gravitational force excerted on me should be the sum of the force excerted by the Earth plus anything else, especially the moon. With the sum of these two vectors being what I feel as "down", should the direction of down change as the moon's position in the sky changes?
I thought I'd do a quick calculation and hear if you guys think it makes sense.
Approximate average distance from the surface of the Earth to the center of the moon is
378 000 km. It has a mass of 7.34*10^22 kg.
The gravitational constant is G = (6.67*10^-11)*(m^3)*(kg^-1)*(s^(- 2))
Which should give me the acceleartion:
a_moon = G * Moon_mass / distance to moon^2
a_moon = (6.67*10^-11)*(m^3)*(kg^-1)*(s^(- 2)) * 7.34*10^22 kg / ((3.78*10^8 m)^2)
a_moon = 3.426x10^-5 m/s^2
With Earth's acceleration at 9.8 and the moon at the horizon we can do some quick trigonometry on the vectors. With the two vectors and their sum a_moon_vector + a_earth_vector forming a triangle, with the sum equal to the hypotenuse, we get an angle A which is the angle between the direction of felt down and the direction towards the centre of the earth.
Tan A = a_moon/a_earth
A = Tan^-1 ( 3.42*10^-5 / 9.8) = ca 2*10^-4 degrees
Is this enough to measured? Since we've got the ratio of the lengths in the triangle, we quickly find that if we have a pendulum hanging from a 300 meter long string (say, from the eiffel tower) it will deviate by 1 millimeter. That is, a change in 2 millimeters as the moon passes from one side of the horizon to the next.