Gravitation Concept Page - 3

Definition

Weight of Object on Moon

The moon is 1/4 the size of Earth and its mass is 1.2×10−2$1.2\times {10}^{-2}$ the mass of earth. Hence, the moon's gravity is much less than the earth's gravity.
gmoon=gearth/6$g_{moon}=g_{earth}/6$
Since weight = mg, weight of an object on moon's surface is 1/6th the weight of the object on earth.

Definition

Variation in acceleration due to gravity

Value of acceleration due to gravity changes with height and depth from the earth's surface. It is maximum on the earth's surface. Its variation with height (or depth) is shown in the plot.

Result

Approximate variation of acceleration due to gravity with height and depth

At a height h<<Re$h<<R_e$, approximate expression for acceleration due to gravity is given by:
g′=g(1−2hRe)$g^{\prime }=g(1-{\displaystyle \frac{2h}{R_e}})$

At a depth d$d$, expression for acceleration due to gravity is given by:
g′=g(1−dRe)$g^{\prime }=g(1-{\displaystyle \frac{d}{R_e}})$

Definition

Effect of non-spherical shape of earth on the value of acceleration due to gravity

Earth is not perfectly spherical in shape. It is flattened at poles and bulging at the equator. Hence, value of acceleration due to gravity is more on the poles than at the equator. 

Example

Affect of sun's gravitation on earth's surface

Example: The masses of sun and earth are Ms$M_s$ and Me$M_e$, respectively and the distance between them is Res$R_{es}$ then, what is the distance of a body of mass m, from the earth along the line towards the sun, where the sun's
gravitational pull balances that of the earth?

Solution:The net gravitational pull is zero i.e. Fs=Fe....................(1)$F_s=F_e....................(1)$
Let, the distance from earth is x$x$. So, the distance from sun is Res−x$R_{es}-x$
Now,Fs=GMsm(Res−x)2$F_s={\displaystyle \frac{G{M}_{s}m}{(R_{es}-x{)}^2}}$ and
Fe=GMemx2$F_e={\displaystyle \frac{G{M}_{e}m}{x^2}}$.
Substituting in (1)$(1)$ we get,
GMsm(Res−x)2=Fe=GMemx2${\displaystyle \frac{G{M}_{s}m}{(R_{es}-x{)}^2}}=F_e={\displaystyle \frac{G{M}_{e}m}{x^2}}$
Msm(Res−x)2=Memx2${\displaystyle \frac{M_{s}m}{(R_{es}-x{)}^2}}={\displaystyle \frac{M_{e}m}{x^2}}$
On further simplication, x=ResMsMe−−−−√+1$x={\displaystyle \frac{R_{es}}{\sqrt{{\displaystyle \frac{M_s}{M_e}}}+1}}$

Example

Variation of acceleration due to gravity with latitudinal position on earth

g2=g2(1−2ω2Rcos2φg)$g^2={\displaystyle g^2(1-\frac{2{\omega }^{2}Rco{s}^2\phi }{g})}$
g=g(1−2ω2Rcos2φg)12$g={\displaystyle g(1-\frac{2{\omega }^{2}Rco{s}^2\phi }{g})\frac{1}{2}}$
Applying binomial expansion and neglecting higher powers we get,
g=g(1−12×2ω2Rcos2φg)$g={\displaystyle g(1-\frac{1}{2}\times \frac{2{\omega }^{2}Rco{s}^2\phi }{g})}$
g=g(1−ωRcos2φg)$g={\displaystyle g(1-\frac{{\omega }^{R}co{s}^2\phi }{g})}$
g=g−ω2Rcos2φ$g=g-{\omega }^{2}Rco{s}^2\phi$
This gives the acceleration due to gravity due to rotation of earth.
At equator; φ=0$\phi =0$
g=g−ω2R$g=g-{\omega }^{2}R$
At a latitudinal position ϕ$\varphi$ on earth, where w$w$ is rotational velocity of the earth, the variation on gravitational acceleration follow above relationship.

Example

Problems on relation of g and G

Example:Calculate the mass of Earth when acceleration due to gravity on earth surface is 9.8 m/s2$9.8m/s^2$.
Solution :From the expression g=GMR2$g={\displaystyle \frac{GM}{R^2}}$ ,
The mass of the Earth can be calculated as follows: gR2=GM$g{R}^2=GM$
M=9.8×(6.38×106)26.67×10−11$M={\displaystyle \frac{9.8\times (6.38\times {10}^6{)}^2}{6.67\times {10}^{-11}}}$
M=5.98×1024 kg$M=5.98\times {10}^{24}kg$.

Definition

Gravitational field lines

Gravitational field lines converge to masses and form equipotential surfaces in equal radial distances.

Definition

Gravitational Field

A gravitational field is a model used to explain the influence that a massive body extends into the space around itself, producing a force on another massive body.
Gravitational field has units N/kg.

Example

Gravitational force due to continuous masses by integration of elemental masses

Example: A uniform ring of mass m$m$ is lying at a distance 3√$\sqrt{3}$a$a$ from the centre of a sphere of a mass M just over the sphere (where a$a$ is the radius of the ring as well as that of the sphere). Then what is magnitude of gravitational force between them?
Solution:Net force on ring=∫wholeringdFsin60∘$={\int }_{wholering}dF\sin {60}^{\circ }$
   =∫wholeringGM(dm)(2a)23√2${\displaystyle ={\int }_{wholering}\frac{GM\left(dm\right)}{{\left(2a\right)}^2}\frac{\sqrt{3}}{2}}$
   =3√GMm8a2${\displaystyle =\frac{\sqrt{3}GMm}{8a^2}}$
as ∫wholeringdm=m${\int }_{wholering}dm=m$

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