Equilibrium and Pressure Distribution in a Spherical Body Under Gravity

In summary, the pressure at r due to the mass on top is proportional to the mass on top multiplied by the gravitational acceleration at that radius.
  • #1
erisedk
374
7

Homework Statement


A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own gravity. If P(r) is the pressure at r (r<R) then the correct option(s) is(are): (more than one correct)
(A) P(r = 0) = 0
(B) [P(r=3R/4)] / P(r=2R/3) = 63 / 80
(C) [P(r=3R/5)] / P(r=2R/5) = 16 / 21
(D) [P(r=R/2)] / P(r=R/3) = 20 / 27

Homework Equations

The Attempt at a Solution


I believe this is a question from gravitational pressure, but I don't really understand how to solve problems on this concept. I know that you have to take an element of thickness dr at a distance r from the centre. And do something with that. But I don't understand anything about this concept. Please help?
 
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  • #2
Then you need to revise your course notes on gravitational pressure.
What causes it? What would you expect: would the pressure increase or decrease as you go towards the center? Why?
What would the amount of pressure depend on?
 
  • #3
Consider an element at radius s, r<s<R, thickness ds. What is the gravitational acceleration at radius s? What weight has the element? What does it add to the pressure at radius r?
 
  • #4
Simon Bridge said:
Then you need to revise your course notes on gravitational pressure.
What causes it? What would you expect: would the pressure increase or decrease as you go towards the center? Why?
What would the amount of pressure depend on?

What causes it? Gravitational force.
What would you expect: would the pressure increase or decrease as you go towards the centre? Increase.
Why? Because we're putting more mass on top of the elemental shell, so the pressure would increase.
What would the amount of pressure depend on? P=F/A = m(on top of it, kind of like in Gauss's law we only consider charge outside the Gaussian surface) / 4πs2 where s is the radius of the elemental shell we're considering.
 
  • #5
haruspex said:
Consider an element at radius s, r<s<R, thickness ds. What is the gravitational acceleration at radius s? What weight has the element? What does it add to the pressure at radius r?

What is the gravitational acceleration at radius s? Gravitational field is (GM/R3).s
What weight has the element? dm = ρ4πs2ds.
What does it add to the pressure at radius s? Since P=F/A = (dm)g/A = { ρ4πs2ds . GM/R3 . s } / 4πs2
P(due to all the mass on top) = GMρ/R3 ∫ sds (from s=s to s=R)
= GMρ/2R { 1-s2/R2 }

I got the answer and I finally understand this concept! Thanks :)
 
  • #6
Well done.
 

What is gravitational pressure?

Gravitational pressure is the force exerted by a massive body on another body due to the gravitational attraction between them. It is responsible for the movement of objects towards the center of the Earth.

How is gravitational pressure calculated?

Gravitational pressure is calculated using the formula P = ρgh, where P is the pressure, ρ is the density, g is the acceleration due to gravity, and h is the height or depth of the object.

What factors affect gravitational pressure?

Gravitational pressure is affected by the mass and distance between two objects. The greater the mass of an object, the stronger the gravitational pressure will be. The further apart two objects are, the weaker the gravitational pressure will be.

How does gravitational pressure impact Earth and other planets?

Gravitational pressure is responsible for the formation and maintenance of planetary systems. It keeps planets in orbit around the sun and holds moons in orbit around planets. It also plays a role in shaping the structure and composition of a planet.

Can gravitational pressure be negative?

No, gravitational pressure cannot be negative. It is always a positive value, meaning it always acts towards the center of the object exerting the force. However, the net pressure on an object may be negative if there are other forces acting on it in the opposite direction.

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