9—Vector Calculus 1 2789.25 A fluid of possibly non-uniform mass density is in equilibrium in a possibly non-uniform gravitational field.
Pick a volume and write down the total force vector on the fluid in that volume; the things acting on it are
gravity and the surrounding fluid. Take the limit as the volume shrinks to zero, and use the result of the preceding
problem in order to get the equation for equilibrium.
(b) Apply the result to the special case of a uniform gravitational field and a constant mass density to find the
pressure variation with height. Starting from an atmospheric pressure of 1. 01 × 105 N/m^2 , how far must you go
under water to reach double this pressure? Ans: about 10 meters
9.26 The energy density,u=dU/dV, in the gravitational field isg^2 / 8 πG. Use the results found in Eq. ( 35 ) for
the gravitational field of a spherical mass and get the energy density. An extension of Newton’s theory of gravity
is that the source of gravity isenergynot just mass! This energy that you just computed from the gravitational
field is then the source of more gravity, and this energy density contributes as a mass densityρ=u/c^2 would.
Find the additional gravitational fieldgr(r)that this provides and add it to the previous result forgr(r).
(b) For our sun, its mass is 2 × 1030 kg and its radius is 700 , 000 km. Assume its density is constant throughout
so that you can apply the results of this problem. At the sun’s surface, what is the ratio of this correction to the
original value?
(c) What radius would the sun have to be so that this correction is equal to the originalgr(R), resulting in double
gravity?
9.27 Continuing the ideas of the preceding problem, the energy density,u=dU/dV, in the gravitational field is
g^2 / 8 πG, and the source of gravity is energy not just mass. In the region of space that is empty of matter, show
that the divergence equation for gravity, ( 32 ), then becomes
∇.~g=− 4 πGu/c^2 =−g^2 / 2 c^2Assume that you have a spherically symmetric system,~g =gr(r)ˆr, and write the differential equation forgr.
Solve it and apply the boundary condition that asr→∞, the gravitational field should go togr(r)→−GM/r^2.
How does this solution behave asr→ 0 and compare its behavior to that of the usual gravitational field of a
point mass.
(b) Can you explain why the behavior is different? Note that in this problem it’s the gravitational field itself that
is the source of the gravitational field; mass as such isn’t present.
(c) A characteristic length appears in this calculation. Evaluate it for the sun.
Ans: (a)gr=−GM
/[
r(r+R)]
, whereR=GM/ 2 c^2