Ah he seems to get confused with his terminology - he shows Pin and Gravity first as being the forces causing compression - he then discontinues using Pin and starts uses Pinside and I think he means that Pinside is the pressure being exerted by the star to counter Pin.
The Newtonian equation for gravity could be expressed as Gm(r)/r^2 where m(r) is the mass contained within a specific radius (e.g. if you wanted to calculate the gravity half way between the surface and the centre of the Earth, you would only allow for the mass within that specific radius). This of course is relative to density and interestingly, Earths gravity increases first before it begins to decrease as you travel from the surface to centre, eventually reaching zero at the centre.
Gravity wouldn't be negative at the centre as m(r) reduces to zero at the centre. If you look at gravity just outside of the outer core, the radius from the centre is ~3.85x10^6 m and the mass contained within this radius is ~39% the total mass of the Earth, g=Gm(r)/r^2 = G(6x10^24 x 0.39)/(3.85x10^6)^2 = 10.6 m/s^2, higher than the surface, but if you look at gravity just outside the inner core (m(r)=1.8% Earth mass, r=1.638x10^6) the gravity reduces to ~2.74 m/s^2.
@nutytube This is because an object with mass has a "centre of gravity". The point at which gravity appears to act. Therefore, when taking gravity into affect, you would commonly use the centre of a star because stars are (give or take a bit of mass) perfectly spherical meaning the "centre of gravity" of a star is at it's centre.
@nutytube This is because an object with mass has a "centre of gravity". The point at which gravity appears to act. Therefore, when taking gravity into affect, you would commonly use the centre of a star because stars are (give or take a bit of mass) perfectly spherical meaning the "centre of gravity" of a star is at it's centre.
Ah he seems to get confused with his terminology - he shows Pin and Gravity first as being the forces causing compression - he then discontinues using Pin and starts uses Pinside and I think he means that Pinside is the pressure being exerted by the star to counter Pin.
okeeft01 2 years ago
Hang on, (Pin and Gravity are the the forces causing compression and Pout is resisting that compression)
Shouldn't Pin + G = Pout at Equilibrium, therefore (remembering my basic algebra)
Pin - Pout = - G
and Pout - Pin = G, he was right first time at 2.34?
okeeft01 2 years ago
No, he's asssuming Pin and Pout are vectors, i which case he's right the second time.
frognyanya 2 years ago
I see.
nutytube 3 years ago
my biggest issue with gravity is why does it seem to concentrate in the middle of the mass?
despite the fact that the mass around the point in the absolute middle is what creates gravity, so how does that come together?
nutytube 3 years ago
The Newtonian equation for gravity could be expressed as Gm(r)/r^2 where m(r) is the mass contained within a specific radius (e.g. if you wanted to calculate the gravity half way between the surface and the centre of the Earth, you would only allow for the mass within that specific radius). This of course is relative to density and interestingly, Earths gravity increases first before it begins to decrease as you travel from the surface to centre, eventually reaching zero at the centre.
stevebd1 3 years ago
thank you very much for the reply!
but what causes this point of swing?
nutytube 3 years ago
and is there any chance it can be negative in the middle?
nutytube 3 years ago
Gravity wouldn't be negative at the centre as m(r) reduces to zero at the centre. If you look at gravity just outside of the outer core, the radius from the centre is ~3.85x10^6 m and the mass contained within this radius is ~39% the total mass of the Earth, g=Gm(r)/r^2 = G(6x10^24 x 0.39)/(3.85x10^6)^2 = 10.6 m/s^2, higher than the surface, but if you look at gravity just outside the inner core (m(r)=1.8% Earth mass, r=1.638x10^6) the gravity reduces to ~2.74 m/s^2.
stevebd1 3 years ago
I see.
nutytube 3 years ago
@nutytube This is because an object with mass has a "centre of gravity". The point at which gravity appears to act. Therefore, when taking gravity into affect, you would commonly use the centre of a star because stars are (give or take a bit of mass) perfectly spherical meaning the "centre of gravity" of a star is at it's centre.
master1140 1 year ago
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@nutytube This is because an object with mass has a "centre of gravity". The point at which gravity appears to act. Therefore, when taking gravity into affect, you would commonly use the centre of a star because stars are (give or take a bit of mass) perfectly spherical meaning the "centre of gravity" of a star is at it's centre.
master1140 1 year ago
is there any explain why is the corona has much higher temperature than the surface?
nutytube 3 years ago
very interesting...
I do hope that uni lectures will be a feature of your channel.
nutytube 3 years ago