3
$\begingroup$

The gravitational acceleration due to the Earth at its surface, $g_E$, is $$g_E = 9.8\hbox{ m/s${^2}$}.$$ The gravitational acceleration due to the Sun at the Earth's position is $$g_S = 5.9 \times 10^{-3}\hbox{ m/s${^2}$}.$$ The gravitational acceleration due to the Moon at the Earth's position is $$g_M = 3.3 \times 10^{-5}\hbox{ m/s${^2}$}.$$ Therefore the change in weight of a $70$ kg person due to the daily change of the Sun's direction, $\Delta W_S$, should be roughly $$\Delta W_S=70\hbox{ kg} \times \frac{g_S}{g_E}=42\hbox{ g}.$$ Additionally the change in weight of a $70$ kg person due to the daily change of the Moon's direction, $\Delta W_M$, should be roughly $$\Delta W_M=70\hbox{ kg} \times \frac{g_M}{g_E}=0.24\hbox{ g}.$$ But does the fact that the Earth is in orbit around the Sun (i.e. we are in free-fall with respect to the Sun) mean that there is no effect of the Sun's gravity on weight measurements here on Earth (neglecting tidal effects)?

Recent readers' questions on this topic in the New Scientist

enter image description here

enter image description here

$\endgroup$
3
  • 1
    $\begingroup$ Related: Precision Pendulum Clocks, Gravity and Tides $\endgroup$
    – PM 2Ring
    Commented Mar 14, 2023 at 10:10
  • $\begingroup$ Kilogram is a unit of mass, not weight. Weight is measured in newtons. $\endgroup$
    – Sandejo
    Commented Mar 14, 2023 at 18:52
  • 1
    $\begingroup$ The tidal effects are the effects. The gravity of the other body is "used up" providing the centripetal acceleration of the orbit. The sun and the moon affecting weight measurements is basically what the tides are (along with the centrifugal effect of the orbit itself). $\endgroup$ Commented Mar 15, 2023 at 1:10

2 Answers 2

2
$\begingroup$

As I understand, your question is: at noon, the solar gravity attracts an object on a scale what would turn it lighter. At midnight its gravitational force adds to the Earth's one, and the object would measure a little bit more.

The fail in this reasoning, thinking only in Newtonian mechanics, is that it supposes a static situation where the object is in equilibrium under several forces: $mg_E, mg_S, mg_M$ and the Normal reaction.

But the object is not in equilibrium. It is accelerated by the Sun (together with the Earth). So, that static analysis is not valid.

The only effect comes from tidal forces, that are too small to be measured by a usual scale.

$\endgroup$
1
$\begingroup$

The Earth itself is in free-fall around the Sun. But people (and everything else) on the surface of the Earth are not quite in free-fall around the Sun because (a) the Earth is rotating and (b) things on the surface of the Earth are at slightly different distances from the Sun, so if they were in free-fall their orbital periods would differ from the Earth's and they would move ahead or fall behind the Earth.

The fact that things on the surface of the Earth are constrained to follow non-free-fall paths around the Sun means that they experience forces called tidal forces. The same thing applies to the Moon. The combined tidal forces from the Sun's and the Moon's gravity cause the ocean tides.

$\endgroup$
3
  • 1
    $\begingroup$ Ok - so apart from tidal forces, can we say that, to a first approximation, there is no effect of the Sun's gravity on Earth due to the fact that we are in free-fall around the Sun? $\endgroup$ Commented Mar 14, 2023 at 10:26
  • 1
    $\begingroup$ @JohnEastmond The Sun's gravity obviously keeps the Earth and everything on it in orbit around the Sun. Solar tidal forces are the residual differences in the Sun's gravitational effect at different locations on the Earth's surface. $\endgroup$
    – gandalf61
    Commented Mar 14, 2023 at 11:30
  • $\begingroup$ @JohnEastmond You're question is basically "how do tidal effects affect g on Earth", so you can't say "apart from tidal effects". You may want to analyze in a rotating reference frame to include centrifugal forces, which is one way to explain the bulge on the far side. $\endgroup$
    – JEB
    Commented Mar 14, 2023 at 15:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.