[geocentrism] Re: The Sun/ Comment & question
- From: "Philip" <joyphil@xxxxxxxxxxx>
- To: <geocentrism@xxxxxxxxxxxxx>
- Date: Mon, 9 Aug 2004 10:42:02 +1000
You can use the same expression to weigh the earth, knowing the
distance of the moon and the time it takes to orbit.
Alan Griffin
It seems you would need to know the G constant for the sun. How would you
obtain that?
I'm no good at big maths Alan Could you please calculate for me what the
weight of this moon would be if the earth was indeed stationary,, and the
moon orbited us once in 24 hours, less that 28 day slip of course if it
mattered much.
Philip.
----- Original Message -----
From: "Alan Griffin" <ajg@xxxxxxxxxxxxx>
To: <geocentrism@xxxxxxxxxxxxx>
Sent: Monday, August 09, 2004 5:26 AM
Subject: [geocentrism] Re: The Sun/ Comment & question
On 07 Aug, Philip <joyphil@xxxxxxxxxxx> wrote:
> Thanks for that Alan. Another similar question. How do they calculate
> the mass of the sun with any degree of accuracy. And the earth for that
> matter, as the composition of the interior is only presumed, as indeed
> all the heavenly bodies.
The mass of a body is easily calculated by timing a satellite
going round it, and knowing its distance.
The mass of the sun in Kg = (Radius of orbit in metres)cubed x
(angular velocity of earth in radians/sec)squared/Gravitational constant G)
= (1.5x10^11)^3 x (2xpi/(364.25 x 24 x 3600))^2 / 6.67x10^-11
which gives 2 x 10^30 kg or 2 x 10^27 tonnes!
You can use the same expression to weigh the earth, knowing the
distance of the moon and the time it takes to orbit.
Alan Griffin
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