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At noon the sun and the earth pull the objects on the earth's surface in opposite directions. At midnight, the sun and the earth pull these objects in same direction. Is the weight of an object, as measured by a spring balance on the earth's surface more at midnight as compared to it's weight at noon? As if I draw the gravitational force vectors of sun & earth it seems that the statement must be true. Also does the answer is affected if I use equal-beam balance instead of spring balance.

Qmechanic
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Avi
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2 Answers2

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An average man weighs around $70kg$.

The force which the earth applies on the man is approximately $685\pm 15N$ $(mg)$.

The force which the sun applies on the man can be calculated as follows:

distance between the sun and the earth $= 1.5 \times 10^{11} m$

mass of the sun $= 2 \times 10^{30}kg$

$$F = G\frac{Mm}{r^2} \approx 0.5N$$

This $0.5N$ is the force that keeps the man in the Earth's orbit around the sun. The earth is revolving around the sun and so is the man. If the man has to rotate in a circular orbit around the sun (the earth's orbit is not circular but it that isn't important here), the sun must apply a centripetal force. This centripetal force cannot be measured by your spring balance.

However, as you state, you are at opposite ends of the earth during the day and night. The difference in Sun's gravitational pull is negligible. The diameter of the earth is just a little over $1.2 \times 10^7 m$. That is 10,000 times smaller than the distance between the earth and the sun. That will barely have any effect on the weight of the man.

If you are interested, read about tidal forces due to the sun. The difference in gravitational forces due to the sun/moon at different ends of the earth are responsible for the high tides and low tides. They play a significant role in ocean waves but the tidal force has no real significance for a man's weight.

You should be more concerned about the variation of gravity over the earth's uneven surface or perhaps, the gravitational force applied on the man by the moon.

Yashas
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Assuming that the gravitational force of the sun does not change much due to the increase in distance (a good approximation), then the sun does not influence the weight because Earth, the balance and the person are already on free fall around the sun. For instance, if earth's gravity were zero, a person would feel weightless, he will not "feel" attracted to the sun.

To make this argument more explicit: Imagine earth in a circular motion of radius $R$ around the sun. At noon the forces on the person will be:

$F_{sun}+N_{noon}-F_{earth}=m\omega^2(R-r)$,

where r is the radius of earth, $\omega$ the angular speed around the sun, and N the normal force from the balance (that is, the measured weight). At midnight the expression above changes to:

$F_{sun}-N_{midnight}+F_{earth}=m\omega^2(R+r)$

From them you obtain:

$N_{midnight}-N_{noon}=2F_{sun}-2m\omega^2r$

Using $F_{sun}=m\omega^2R$, we get:

$N_{midnight}=N_{noon}$