If a pendulum clock keeps perfect time at the base of a mountain, will it also keep perfect time when moved to the top of a mountain?
I know that as the value of g is lowered, the period of a pendulum becomes longer, so is the value of g lower than 9.8 m/s^2 on top of a mountain?
2006-12-18 07:53:54 · 6 answers · asked by beautyqueenjustine 3 in Science & Mathematics ➔ Physics
No, gravity is slightly less at a higher elevation.
2006-12-18 07:56:31 · answer #1 · answered by computerguy103 6 · 0⤊ 0⤋
I disagree with jj's answer. It's correct that General Relativity predicts that time passes very slightly faster at the top of a mountain than at sea level, and it's correct that gravity is less at the top of the mountain, so the pendulum will swing more slowly and the clock will run slow. Both these effects will work the same way; the clock will fall behind. The General Relativity effect is much, much less than the slowing of the pendulum, and negligible by comparison. But time can be measured more accurately than any other physical quantity; to about 17 decimal places. The Global Positioning System works by timing signals from satellites. It's so accurate that they had to include both Special and General Relativity corrections to allow for the effect on time of the satellite's velocity and its height above the ground.
2006-12-18 08:36:25 · answer #2 · answered by zee_prime 6 · 1⤊ 0⤋
The value of g is indeed lower on top of a mountain. You can prove this by looking at the equation for the inverse square law solved for g.
This equation is g=(G*Mass of earth)/radius^2
If you increase the radius between where you are and the center of the earth, you will get a lower value for g.
Since the period of a pendulum is dependent on the value of g, as you already know, the pendulum clock will not keep perfect time on top of a mountain.
2006-12-18 08:33:03 · answer #3 · answered by ruadhdarragh 3 · 1⤊ 0⤋
It is lower but not by enough to make a noticable change in the amount of time that the pendulum swings. Theoretically the amount of time the pendulum takes is slower, but it could only be measured over a long sumation of time.
2006-12-18 08:01:22 · answer #4 · answered by Jeremy W 2 · 0⤊ 0⤋
Interesting question; gravity has slightly less effect but time moves very slightly slower at higher elevation from the earth (relativity and all that) so maybe the effects cancel each other out; I suppose it depends on the height of the mountain :-)
2006-12-18 08:13:34 · answer #5 · answered by jj 2 · 0⤊ 1⤋
At sea point the fee of g is given by making use of g0 = GM/R^2 above sea point use the formulation g = GM/(R+h)^2 separate out the top version g = GM/R^2 * (a million+h/R)^-2 which for small h/R is g = g0 * (a million-2h/R) g = g0 - 2*g0/R *h g =g0 - 0.003073* h(km) subsequently the consequence of top is to shrink the fee of g by making use of approximately 0.003 for each 1km in top. The results of rotation on the effectual rotation is R* omega^2 * cos(phi) the place omega is the angular rotation in radians in step with 2d = 2*pi(24*60*60) phi = perspective of selection. yet this additionally impacts sea point so as that the fee of R at sea point on the pole is far less that the fee of R on the equator
2016-10-15 04:43:48 · answer #6 · answered by ? 4 · 0⤊ 0⤋