Legend has it that observing an apple falling
from a tree, Newton was inspired to arrive at an
universal law of gravitation that led to an
explanation of terrestrial gravitation as well as
of Kepler’s laws. Newton’s reasoning was that
the moon revolving in an orbit of radius $R_m$ was
subject to a centripetal acceleration due to
earth’s gravity of magnitude
$a_m=\frac{V^2}{R_m}=\frac{4\pi^2 R_m}{T^2}$ (1)
where V is the speed of the moon related to the
time period T by the relation $V=\frac{2\pi R_m}{T}$. The
time period T is about 27.3 days and Rm was
already known then to be about 3.84 × 10$^8$ m. If
we substitute these numbers in Eq. (1), we
get a value of am much smaller than the value of
acceleration due to gravity g on the surface of
the earth, arising also due to earth’s gravitational
attraction.
This clearly shows that the force due to
earth’s gravity decreases with distance. If one
assumes that the gravitational force due to the
earth decreases in proportion to the inverse
square of the distance from the centre of the
earth, we will have $\frac{a_m\alpha}{R_m^2}$ ; $\frac{a_m\alpha}{R_E^2}$ and we get
$\frac{g}{a_m}=\frac{R_m^2}{R_E^2}\simeq 3600$ (2)
in agreement with a value of g 9.8 m s-2 and
the value of am from Eq. (1). These observations
led Newton to propose the following Universal Law
of Gravitation :
Every body in the universe attracts every other
body with a force which is directly proportional
to the product of their masses and inversely
proportional to the square of the distance
between them.
The quotation is essentially from Newton’s
famous treatise called ‘Mathematical Principles
of Natural Philosophy’ (Principia for short).
Stated Mathematically, Newton’s gravitation
law reads: The force F on a point mass $m_2$ due
to another point mass $m_1$ has the magnitude
$|\mathbf{F}|=G\frac{m_1m_2}{r^2}$ (3)
Equation (3) canbe expressed in vector form as
$\mathbf{F}=G\frac{m_1m_2}{r^2}(-\mathbf{\hat{r}})=-G\frac{m_1m_2}{r^2}\mathbf{\hat{r}}$
$\mathbf{F}=-G\frac{m_1m_2}{|\mathbf{\hat{r}}|^3}\mathbf{\hat{r}}$
where G is the universal gravitational constant,
$\mathbf{\hat{r}}$ is the unit vector from $m_1$ to $m_2$ and $\mathbf{r} = \mathbf{r_2} – \mathbf{r_1}$ as shown in Fig. 1.
Fig. 8.3 Gravitational force on $m_1$ due to $m_2$ is along
r where the vector r is ($\mathbf{r_2} – \mathbf{r_1}$).
The gravitational force is attractive, i.e., the
force F is along – r. The force on point mass $m_1$ due to $m_1$ is of course – F by Newton’s third law.
Thus, the gravitational force $\mathbf{F_{12}}$ on the body 1
due to 2 and $\mathbf{F_{21}}$ on the body 2 due to 1 are related
as $\mathbf{F_{12}}$ = –$\mathbf{F_{21}}$.
Before we can apply Eq. (3) to objects under
consideration, we have to be careful since the
law refers to point masses whereas we deal with
extended objects which have finite size. If we have
a collection of point masses, the force on any
one of them is the vector sum of the gravitational
forces exerted by the other point masses as
shown in Fig 2.
Fig. 2 Gravitational force on point mass $m_1$ is the
vector sum of the gravitational forces exerted
by $m_2$, $m_3$ and $m_4$.
The total force on $m_1$ is
$\mathbf{F}=G\frac{m_2m_1}{r_{21}^2}\mathbf{\hat{r_{21}}}+G\frac{m_3m_1}{r_{31}^2}\mathbf{\hat{r_{31}}}+G\frac{m_4m_1}{r_{41}^2}\mathbf{\hat{r_{41}}}$
Example 1
Three equal masses of m kg
each are fixed at the vertices of an
equilateral triangle ABC.
(a) What is the force acting on a mass 2m
placed at the centroid G of the triangle?
(b) What is the force if the mass at the
vertex A is doubled?
Take AG = BG = CG = 1 m (see Fig.3)
Fig. 3 Three equal masses are placed at the three
vertices of the ∆ ABC. A mass 2m is placed
at the centroid G.
Answer
(a) The angle between GC and the
positive x-axis is 30° and so is the angle between
GB and the negative x-axis. The individual forces
in vector notation are
$\mathbf{F_{GA}}=G\frac{(m)(2m)}{1^2}\mathbf{\hat{j}}$
$\mathbf{F_{GB}}=G\frac{(m)(2m)}{1^2}(-\mathbf{i} \ cos \ 30^0 - \mathbf{\hat{j}} \ sin \ 30^0$
$\mathbf{F_{GC}}=G\frac{(m)(2m)}{1^2}(+\mathbf{i} \ cos \ 30^0 - \mathbf{\hat{j}} \ sin \ 30^0$
From the principle of superposition and the law
of vector addition, the resultant gravitational
force $\mathbf{F_R}$ on (2m) is
$\mathbf{F_R}=\mathbf{F_{GA}}+\mathbf{F_{GB}}+\mathbf{F_{GC}}$
$\mathbf{F_{R}}=2Gm^2 \mathbf{\hat{j}}+2Gm^2(-\mathbf{\hat{i}} \ cos \ 30^0-\mathbf{\hat{j}} \ sin \ 30^0)$
$+ 2Gm^2(\mathbf{\hat{i}} \ cos \ 30^0 - \mathbf{\hat{j}} \ sin \ 30^0)=0$
Alternatively, one expects on the basis of
symmetry that the resultant force ought to be
zero.
(b) Now if the mass at vertex A is doubled
then
For the gravitational force between an extended
object (like the earth) and a point mass, Eq. (8.5) is not
directly applicable. Each point mass in the extended
object will exert a force on the given point mass and
these force will not all be in the same direction. We
have to add up these forces vectorially for all the point
masses in the extended object to get the total force.
This is easily done using calculus. For two special
cases, a simple law results when you do that:
(1) The force of attraction between a hollow
spherical shell of uniform density and a
point mass situated outside is just as if
the entire mass of the shell is
concentrated at the centre of the shell.
Qualitatively this can be understood as
follows: Gravitational forces caused by the
various regions of the shell have components
along the line joining the point mass to the
centre as well as along a direction
prependicular to this line. The components
prependicular to this line cancel out when
summing over all regions of the shell leaving
only a resultant force along the line joining
the point to the centre. The magnitude of
this force works out to be as stated above.
(2) The force of attraction due to a hollow
spherical shell of uniform density, on a
point mass situated inside it is zero.
Qualitatively, we can again understand this
result. Various regions of the spherical shell
attract the point mass inside it in various
directions. These forces cancel each other
completely.
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