Universal Law Of Gravitation And Field Intensity

NEWTON'S LAW OF UNIVERSAL GRAVITATION

Between any two particles of masses m₁ and m₂ at separation r from each other there exist attractive forces ${\overrightarrow{F}}_{AB}$ and ${\overrightarrow{F}}_{BA}$ directed from one body to the other and equal in magnitude which is directly proportional to the product of the masses of the bodies and inversely proportional to the square of the distance between the two.

$F_{AB}=F_{BA}=G\frac{m_1{m}_2}{r^2}$

In vector form $\vec{F}AB=-{\vec{F}}_{BA}$.

Where G is universal gravitational constant. G = 6.67 × 10 ^-11 Nm² / kg²

Dimensional formula of g:

$G=\frac{F{r}^2}{m_1{m}_2}=\frac{\left[ML{T}^{-2}\right]\left[L^2\right]}{\left[M^2\right]}=\left[M^{-1}{L}^3{T}^{-2}\right]$

Gravitation force due to a rod:

A uniform rod of mass M of length $\ell$. We will calculate gravitational force on this rod due to a point particle of mass m placed at a distance r from one of its ends as shown.

To find this we consider a small element of width dx on the rod as shown in figure at a distance x from the point mass in. the mass of this element can be given as

$dm=\frac{M}{\ell }dx$

The force on mass m due to dm is,

$dF=\frac{Gmdm}{x^2}ordF=\frac{GmM}{\ell x^2}dx$

To find the net force on rod we integrate the above expression for the whole rod i.e. from a distance r to r+l,

$F=\int dF=\underset{r}{\overset{r+\ell }{\int }}\frac{GMm}{\ell }\frac{1}{x^2}dxorF=\frac{GMm}{\ell }\left[\frac{1}{r}-\frac{1}{r+\ell }\right]$

Illustration 1: A mass M is split into two parts m and (M-m), which are then separated by a certain distance. What ratio (m/M) maximises the gravitational force between the parts.

Solution: If r is the distance between m and (M-m), the gravitational force will be

F= $G\frac{m\left(M-m\right)}{r^2}=\frac{G}{r^2}\left(mM-m^2\right)$

For F to be maximum $\frac{dF}{dm}=0$ as

M and r are constants

i.e. $\frac{d}{dm}\left[\frac{G}{r^2}\left(mM-m^2\right)\right]=0$

i.e. M – 2m = 0 $\left[As\frac{G}{r^2}\ne 0\right]$

$\Rightarrow$ $\frac{m}{M}=\frac{1}{2}$

GRAVITATIONAL FIELD INTENSITY (I or E)

The intensity of gravitational field at point is the gravitational force exerted on a unit mass kept at that point.

$I=\frac{\overrightarrow{F}}{m_0}$$=-\frac{GM}{r^2}\hat{r}$

Dimensional formula of gravitational field intensity

$g=\frac{F}{m}=\frac{ML{T}^2}{M}=\left[M^{0}L{T}^{-2}\right]$

Illustration 2:

The distance between earth and moon is about $3.8\times 10^5$ km. At what point or points will the gravitational field strength of earth-moon system be zero? Given mass of earth is 81 times the moon's mass.

Solution:

In this problem earth and moon will be treated as point masses as they are far apart and as in case of a point mass $I=GM/r^2$, the field due to both the masses will be zero at infinite. In addition to it, the field will also be zero at a point in between earth and moon where the pull of earth balances the pull of moon. If this point is at a distance x from the earth,

$\frac{G{M}_E}{x^2}=\frac{G{M}_M}{{\left(3.8\times 10^8-x\right)}^2}$ or $\frac{81}{x^2}=\frac{1}{{\left(3.8\times 10^8-x\right)}^2}$ [as M_E = 81 M_M

or $9\left(3.8\times {10}^8-x\right)=x$ or $x=3.42\times {10}^{8}m=3.42\times {10}^{5}km$