Gravitation Concept Page - 6

Definition

Gravitational Potential Energy between two point masses

Definition: Gravitational potential energy is energy an object possesses because of its position in a gravitational field. The most common use of gravitational potential energy is for an object near the surface of the Earth where the gravitational acceleration can be assumed to be constant at about 9.8 m/s2$9.8m/s^2$. Since the zero of gravitational potential energy can be chosen at any point (like the choice of the zero of a coordinate system), the potential energy at a height h above that point is equal to the work which would be required to lift the object to that height with no net change in kinetic energy.

Two masses m1$m_1$ and m2$m_2$ separated at a distance r$r$ has potential energy:
U=Gm1m2r$U={\displaystyle \frac{G{m}_1{m}_2}{r}}$

Example

Gravitational Energy of a System

Example: A planet of mass m1$m_1$ revloves round the sun of mass  m2$m_2$. The distance between the sun and the planet is r$r$. Considering the motion of the sun find the total energy of the system assuming the orbits to be circular.

Solution:Velocity of planet is v=Gm2r−−−−−√$v=\sqrt{{\displaystyle \frac{G{m}_2}{r}}}$
And the total energy is P.E+K.E$P.E+K.E$
⇒−Gm1m2R+12mv2=−Gm1m22R$\Rightarrow {\displaystyle \frac{-G{m}_1{m}_2}{R}}+{\displaystyle \frac{1}{2}}m{v}^2={\displaystyle \frac{-G{m}_1{m}_2}{2R}}$

Example

Work done in assembling a system of masses

Example: Three particles, each of mass 10−2${10}^{-2}$ kg are brought from infinity to the vertices of an equilateral triangle of side 0.1 m$0.1m$, what is the work done?

Solution:
Work done,
W=−U=−(−3Gm2r)=3Gm2r$W=-U=-(-3{\displaystyle \frac{G{m}^2}{r})=3\frac{G{m}^2}{r}}$
W=3×6.67×10−11×(10−2)210−1=2×10−13J$W={\displaystyle \frac{3\times 6.67\times {10}^{-11}\times ({10}^{-2}{)}^2}{{10}^{-1}}=2\times {10}^{-13}J}$

Example

Change in potential energy of a mass in a given potential field

Example: A particle of mass m$m$ is located outside a uniform sphere of mass M$M$ at a distance r$r$ from its centre. Find the potential energy of gravitational interaction of the particle and the sphere.

Solution:
Since for a uniform solid sphere,

VG=−GMr$V_G=-\frac{GM}{r}$  for r$r$

and it is the work done to bring a unit mass from ∞$\mathrm{\infty }$ to distance 'r$r$' from sphere.

For mass 'm$m$'

VG=−GMmr$V_G=-{\displaystyle \frac{GMm}{r}}$ which is the total potential energy of the system.

Example

Work done in changing configuration of a system of masses

Example: 3 identical bodies of mass m$m$ are at the corners of an equilateral triangle of side L$L$. What is the amount of work done in moving them such that the side of the equilateral triangle becomes doubled?

Solution:Initial potential energy, Ui=−3Gm2L$U_i={\displaystyle \frac{-3G{m}^2}{L}}$
Final potential energy, Uf=−3Gm22L$U_f=-{\displaystyle \frac{3G{m}^2}{2L}}$
 Work done, W=Uf−Ui=−3Gm22L−(−3Gm2L)=3Gm22L$W=U_f-U_i=-{\displaystyle \frac{3G{m}^2}{2L}-(-\frac{3G{m}^2}{L})=\frac{3G{m}^2}{2L}}$

Example

Work done in changing configuration of a system of continuous masses

Example: Four masses (each of m$m$) are placed at the vertices of a regular pyramid (triangular base) of side ′a′${}^{\prime }{a}^{\prime }$. Find the work done by the system while
taking them apart so that they form the pyramid of side ′2a′${}^{\prime }2a^{\prime }$.

Solution: The initial gravitational potential is given as
−6Gm2a$-6{\displaystyle \frac{G{m}^2}{a}}$
Final gravitational potential is
−6Gm22a$-6{\displaystyle \frac{G{m}^2}{2a}}$
Change in potential is
−6Gm22a−(−6Gm2a)$-6{\displaystyle \frac{G{m}^2}{2a}}-(-6{\displaystyle \frac{G{m}^2}{a}})$
or
6Gm2a−6Gm22a=6Gm22a=3Gm2a$6{\displaystyle \frac{G{m}^2}{a}}-6{\displaystyle \frac{G{m}^2}{2a}}=6{\displaystyle \frac{G{m}^2}{2a}}=3{\displaystyle \frac{G{m}^2}{a}}$
Thus the external work done would be
−3Gm2a$-3{\displaystyle \frac{G{m}^2}{a}}$

Example

Field due to different arrangements of shells

Example: Two concentric shells of masses M1$M_1$ and M2$M_2$ are having radii r1$r_1$ and r2$r_2$. Find the expression of gravitational field in different regions of shells.

Solution:
A uniform shell has no contribution to the gravitational field inside it from Gauss Law for Gravitation. Using this principle,
gravitational field would be 0 for r<r1$r<r_1$
gravitational field would be GM2r2${\displaystyle \frac{G{M}_2}{r^2}}$ for r1<r<r2$r_1<r<r_2$
gravitational field would be G(M1+M2)r2${\displaystyle \frac{G{(M_1+M}_2)}{r^2}}$ for r>r2$r>r_2$

Definition

Aerial Velocity

Areal velocity (also called sector velocity or sectorial velocity) is the rate at which area is swept out by a particle as it moves along a curve. In the adjoining figure, suppose that a particle moves along the blue curve. At a certain time t, the particle is located at point B, and a short while later, at time t + t, the particle has moved to point C. The area swept out by the particle is the green area in the figure, bounded by the line segments AB and AC and the curve along which the particle moves. The areal velocity equals this area divided by the time interval t in the limit that t becomes vanishingly small.

Definition

Kepler's Law

 The orbit of a planet is an ellipse with the Sun at one of the two foci.

Definition

Kepler's Second Law

A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.

Definition

Kepler's Third Law

The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.

Example: The period of revolution of an earth's satellite close to the surface of earth is 90 minutes. Find the period of another earth's satellite in an orbit at a distance of three times earth's radius from its surface.

Solution:
Time period T2=4π2GMR3$T^2={\displaystyle \frac{4{\pi }^2}{GM}}{R}^3$ 
i.e T2∝R3$T^2\propto R^3$
⇒(T1T2)2=(R1R2)3$\Rightarrow {\left({\displaystyle \frac{T_1}{T_2}}\right)}^2={\left({\displaystyle \frac{R_1}{R_2}}\right)}^3$
here R1=Re$R_1=R_e$ and R2=Re+3Re=4Re$R_2=R_e+3R_e=4R_e$

∴(T1T2)2=(Re4Re)3$\therefore {\left({\displaystyle \frac{T_1}{T_2}}\right)}^2={\left({\displaystyle \frac{R_e}{4R_e}}\right)}^3$

⇒902T22=164$\Rightarrow {\displaystyle \frac{{90}^2}{T_2^2}}={\displaystyle \frac{1}{64}}$
⇒T2=90×641/2=90×8=720$\Rightarrow T_2=90\times {64}^{1/2}=90\times 8=720$ minutes

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