# How much would you weigh if Earth stopped orbiting the Sun? [closed]

I have a curious question.

The reason why astronauts are weightless in space is because they are orbiting Earth. If they suddenly stopped orbiting they would fall directly into Earth. Earth is doing the same around the Sun, so it would fall directly into the Sun if the Earth stopped orbiting the Sun.

If the Earth stopped rotating and orbiting, obviously ignoring the adverse effects of doing this and having Earth hold it's current position in it's orbit, how much would a 150 pound person weigh if they stood on suitable position on Earth.

This question is kinda out there, but it's a thought-provoking question for me and hopefully others.

• "Stopped orbiting the sun" is rather vague. The Earth could fly out away from the sun in a straight line and be considered to stop orbiting the sun. Oct 16, 2014 at 18:00
• I suspect the OP wants us to answer the question what does a person weigh if the Earth magically stops moving at 30 km/s relative to the Sun, but the Sun's gravitation remains in effect. Oct 16, 2014 at 18:04
• I think this question falls squarely in the category of hypothetical questions that are in the 'don't ask' category, and I am flagging it for moderator review. astronomy.stackexchange.com/help/dont-ask Oct 17, 2014 at 10:15
• This question appears to be off-topic because it is a "what if" questions, which is currently defined as being off-topic. Oct 29, 2014 at 15:43

In a sense, this is one of those "If I break the laws of physics, what do the laws of physics say will happen" kinds of questions.

Ignoring that, a 150 pound person would still weigh about 150 pounds if the Earth miraculously stopped orbiting the Sun. Almost all of a person's weight results from resistance to the Earth's gravity field. The tidal gravitational forces from the Moon and Sun are on the order of 10-4 ounces of force.

Note: I'm answering the question from the perspective of "what would a bathroom scale say". A doctor's scale would say a 150 pound person weighs 150 pounds regardless of whether the person is standing on the Moon, the Earth, or a very massive planet. A doctor's scale measures mass. A bathroom scale measures apparent weight.

So what would a bathroom scale say? If the Earth is at one astronomical unit from the Sun, whether the Earth is orbiting at 30 km/s or falling straight toward the Sun doesn't matter one iota. All that matters is that the Earth is one AU from the Sun. In both cases, that the Sun is 1 AU distant makes a 150 pound person weigh about $1.8 \times 10^-4$ ounces less at noon or at midnight than at sunrise or sunset.

The difference between the orbiting Earth and the in-falling Earth is that these tidal forces are more or less constant in the case of the orbiting Earth. They grow ever larger in the case of the in-falling Earth as it gets closer and closer to the Sun.

These tidal forces become quite large by the time the surface of the Earth just touches the surface of the Sun. Now our 150 pound person would weigh only 73.5 pounds per a bathroom scale if he was standing at that point where the surface of the Earth just touches the Sun. He would also weigh about the same at the point on the Earth furthest from the Sun. He would however weigh considerably more, about 188 pounds, if he happens to be at a point where the Sun is on the horizon.

If, on the other hand, some mysterious, magical force makes the Earth stop orbiting the Sun and makes this change permanent (i.e., the Earth stops moving with respect to the Sun), a 150 pound person closest to the Sun would weigh about 1.4 ounces less, while a 150 pound person furthest from the Sun would weigh about 1.4 ounces more. A 150 pound person who sees the Sun at the horizon would still weigh about 150 pounds in this scenario.

• So, the person standing on Earth would still weight about the same. I would imagine the gravitational force of the Sun pulling on the Earth would be greater than Earth's gravity thus making a person weigh a lot more.
– Err
Oct 16, 2014 at 17:23
• OK, so you're saying the earth comes to a complete halt with reference to the Sun, and loses its gravitational pull (eg, it vanishes without a trace), is that correct? I was assuming you meant it would travel in a straight line per Newton's First Law.
– user21
Oct 16, 2014 at 17:50
• @barrycenter - I don't know if your comment was aimed at me. I assumed that Err was asking what a person would weigh if the Earth magically stopped moving at 30 km/s relative to the Sun but then not so magically continued falling sunward (which is exactly what the Earth does all the time). Oct 16, 2014 at 18:00
• @Err - You imagine wrong. Oct 16, 2014 at 18:05
• @DavidHammen - Yes, this is assuming if Earth was held still relative to the Sun. Also, I can accept a response to my comment without you being a dick about it. This does however fully answer my question. Thanks.
– Err
Oct 16, 2014 at 20:37

Weighlessness is a property of gravitational freefall. When you're standing on the surface of the Earth, the weight you feel is the reaction force of the Earth's surface stopping you from falling, i.e., weight is caused by the failure of gravitational freefall.

There is no relevant difference between a round orbit and a radial plunge orbit. They're both completely valid orbits as far as the laws of physics are concerned, and the fact that one will eventually smack you into the Earth does not matter until it actually happens.

If they suddenly stopped orbiting they would fall directly into Earth. Earth is doing the same around the Sun, so it would fall directly into the Sun if the Earth stopped orbiting the Sun.

So if the Earth were suddenly plunging into the Sun as you describe, there would be no discernible difference in weight, barring fairly minor tidal effects when one gets closer to the Sun. Conceptually, it would be a different orbit, rather than a lack of orbiting.

If the Earth stopped rotating and orbiting, obviously ignoring the adverse effects of doing this and having Earth hold it's current position in it's orbit, how much would a 150 pound person weigh if they stood on suitable position on Earth.

The Sun has $GM = 1.3271244\times 10^{20}\,\mathrm{m}^3/\mathrm{s}^2$, so at about $r \approx 1\,\mathrm{AU}$, the Earth holding still would give about $GM/r^2 = 5.9\,\mathrm{mm}/\mathrm{s}^2$, which is about $0.06\%$ of Earth's gravity. So they weigh more by about that proportion if on the far side from the Sun, and less if on the near side. This works out to be about $1.4\,\mathrm{oz}$.

The centrifugal force provided by the Earth's rotation is a few times more significant, up to about $34\,\mathrm{mm}/\mathrm{s}^2$, depending on the location on the Earth.

• This does not answer my question, but I do understand that you are using a technical definition of the word "orbit". Relative to Earth, I understand that one cannot technically stop "orbiting" unless they are standing on Earth already. Poor choice of words on my part, but the meaning of my question is still the same.
– Err
Oct 16, 2014 at 17:19
• Actually, it's a lot, lot less than 1.4 ounces, by many orders of magnitude. The Earth as a whole falls sunward at 5.9 mm/s$^2$. The difference between this and the sunward acceleration at the point on the surface of the Earth closest to the Sun is $5\times10^{-4}$ mm/s$^2$. Multiply that by 150 pounds (mass) and you get about $10^{-4}$ ounces of force. Oct 16, 2014 at 17:55
• @DavidHammen no, what you're calculating are tidal forces. I interpreted the question as comparing the weight when the Earth is held still vs when it's orbiting, not at comparing different locations on the Earth. When it's held still, the Sun's $5.9\,\mathrm{mm}/\mathrm{s}^2$ is felt in addition to Earth's gravity (well, or subtractively, depending on location). Oct 16, 2014 at 18:11