# How much does the sun weigh?

#### marcus

Gold Member
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At this distance from the sun, how much would a copy weigh?

In other words, with what force (in newtons, or planck units, or dynes, or tons---your choice) would a copy be attracted if it were at earth's average distance?

So raise that to the fourth-----10^-16

That 10^-16 is the sun's weight in Planck units (c=G=hbar=1)

Do you believe that gives the correct answer, and if so, why?

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#### marcus

Gold Member
Dearly Missed
dav2008 YES that is the right thing to pick
for planck acceration
two small black holes colliding might attract
each other so much just before they merge
that they accelerate towards each other with
such extreme acceleration, I think, or at least
something approaching that.

what would you say the sun weighs, in newtons?

planck force is this huge force of 12E43 newtons
and so I am predicting E-16 of that
can you verify it?

#### dav2008

Gold Member
Originally posted by marcus

what would you say the sun weighs, in newtons?

Hmm? Im kind of confused with respect to what are u measuring the sun's weight? It has a definite mass...u can look it up somewhere..

Are u replying to the other topic and posted a new one accidentally? Or is this a new one related to that?? *Confused*

#### marcus

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Originally posted by dav2008
Hmm? Im kind of confused with respect to what are u measuring the sun's weight? It has a definite mass...u can look it up somewhere..

Are u replying to the other topic and posted a new one accidentally? Or is this a new one related to that?? *Confused*
sorry I caused you confusion---I was doing two things in one post. First, what you said about planck acceleration is absolutely right! It is unit speed over unit time: c over planckunittime.

And just like in metric, where unit force (newton) is what gives
a unit of acceleration to the unit mass,
in planck the unit force (planck force) is what gives
a planck mass the unit acceleration.
it turns out to be c^4/G and in metric terms it is 12E43 newtons.

this other question about the sun's weight is a separate issue
With what force in newtons would the sun attract a copy of itself placed at this distance----150 million kilometers.

In other words what would the sun weigh in a gravitational field like that which the earth experiences from the sun.

In planck terms this force (which would be the sun's weight-force at this distance) is very simple to calculate. It is just 10^-16 of the universal force constant or planck force. But you can also calculate if from the masses of the sun and its copy expressed in kilograms, and the metric value of G = 6.673x10^-11 m^3s^-2 per kilogram, and the distance between them by the usual law
that the force is GMm/r^2.

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#### marcus

Gold Member
Dearly Missed
Originally posted by marcus
At this distance from the sun, how much would a copy weigh?

In other words, with what force (in newtons, or planck units, or dynes, or tons---your choice) would a copy be attracted if it were at earth's average distance?

So raise that to the fourth-----10^-16

That 10^-16 is the sun's weight in Planck units (c=G=hbar=1)

Do you believe that gives the correct answer, and if so, why?
dav commented but no one responded with a calculation, so I will reply to my own questions in this thread:

Working with c=G=hbar=k=1 units one of the few data I remember are the earth's orbit speed E-4 (one tenthousandth of the speed of light) and distance to the sun 93E44 (93 million miles)

From that, by squaring the speed (E-8) and multiplying that by the distance I can tell that the mass of the sun is 93E36

And by squaring the square of the speed (E-16) I can tell that the force of the sun's weight would be E-16 at this distance from a twin. Two copies of the sun placed 93 million miles apart would attract each other with that force.

Check: To check that the weight is E-16, square the mass (93E36^2) and divide by the square of the distance (93E44^2).
You get E72 divided by E88 which indeed is E-16.

In metric terms the planck force unit is 12E43 newtons, so the force E-16 is 12E27 newtons. If you want a further check, calculate the sun's weight at this distance entirely in metric terms and it will come out to be 12E27 newtons.

Sun's mass = 2e30 kg
Average distance = 1 AU = 1.496e11 meters

a = GM/r2
F = ma
go nuts.

#### marcus

Gold Member
Dearly Missed
Originally posted by marcus
At this distance from the sun, how much would a copy weigh?

In other words, with what force (in newtons, or planck units, or dynes, or tons---your choice) would a copy be attracted if it were at earth's average distance?

So raise that to the fourth-----10^-16

That 10^-16 is the sun's weight in Planck units (c=G=hbar=1)

Do you believe that gives the correct answer, and if so, why?
I sympathise with no one wanting to work the weight out in newtons. It is easier to work out in planck-----the force is in fact 10-16
the fourth power of the circ. orbit speed at that distance.

What's the moon's weight in its own surface gravity?
the speed of a skimming orbit is 5.6x10-6

(5.6 millionths of the speed of light, or 5.6 times a common
speed for sound at aircraft cruising altitudes)

Raising 5.6x10-6 to the fourth gives 10-22 which is the moon's weight in planck

I also would not want to bother working it out in metric (you aren't the only one!) but the answer in metric is easy to give because planck force is 12E43 newtons so that the moon's weight, being E-22 of that, is evidently 12E21 newtons.

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