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8/8/2019 Newton Shell
1/2
Newtons shell theorem
From Physclips: www.physclips.unsw.edu.au
This theorem shows that a spherically symmetric body, mass Mbody, exerts a gravitational force on an
external object, mass m, that equals the force between m and a point mass Mbody, located at the centre of
the spherical distribution. Many astronomical objects (especially stars and planets) are nearly sphericallysymmetric. Consequently, we can calculate the gravitational forces between them using their masses and
the distance between their centres. Lets see why.
Consider first a thin sphericallysymmetric shell (dark shading) ofmass M and radius R and a pointmass m at r from its centre, asshown. First we consider the case r> R. We need to integrate thegravitational attraction between mand all parts of M.
For the integration, we slice theshell up into narrow rings (lightshading), so that all points in thering are the same distance, s, fromm. Let the ring subtend an angle d,as shown.
d
R
dM
r
M Fgrav
m
s
We define , the area density or mass per unit area of the shell: =M
4R2 . So the mass of the ring is dM
is times its area. Its circumference is 2Rsin and its width is Rd, so
dM = .2Rsin.Rd =Msin.d
2
The force exerted by each part of the ring is a vector, and we must add these vectors. From symmetry,however, the gravitational force on m will be along the axis of the ring, so we need only add the vectors onthe axis. Each tiny part along the perimeter of the ring exerts a force towards it, and the axial component ofeach force includes a factor cos.
dF =GmdM
s2cos =
GmM
2s2sin cos d
From the cosine rule, we can write
cos =r2+R2s2
2rR and cos =r2+s2R2
2rs
All three of, and s vary. Let's make s the independent variable. Differentiating both sides of theequation for cos gives
sin d =s
rR ds
We can then subsitute this equation, plus the expression for cos in our expression for dF
dF =GmM
2s2
r2+s2R22rs
srR ds =
GmM
4Rr2
r2R2+s2s2 ds
8/8/2019 Newton Shell
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To include the whole shell, s varies from (rR) to (r+R), so the force due to the whole shell is
F =GmM
4Rr2
s=rR
r+R r2R2+s2s2
dsThe integrand as the form (constant2 + s2)/s2, so the integral is a standard one. The definite integral has the
value 4R, so we obtain the simple expression
F =GmM
r2
For a spherically symmetric distribution of mass, we can then integrate over R. Each shell gives a similarexpression.
Now look at the case where r < R,i.e. when m is inside the hollowshell. In this case, integrating overthe sphere requires that s go fromR-r to R+r: this changes the sign of
the lower limit of integration. Inthis case, the integral gives zero. So,inside a hollow shell, the totalgravitational field due to the shell iszero.
This may seem odd: in thisdiagram, most of the shell is to theleft of m. However, some of theparts to the right of m are closer tom, and these effects cancel out.
d
R
dM
r
M
m
s
Combining the two results, we see that, at a point inside a symmetric distribution of mass, only the mass closerto the centre contributes to the gravitational field.
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