Gravitation Concept Page - 8

Formula

Escape Velocity on moon

Consider a satellite of mass m, stationary on the surface of moon, then binding of the satellite on the surface of moon is given by
B.E.=GMmmR$B.E.={\displaystyle \frac{G{M}_{m}m}{R}}$
R- radius of moon
To escape from the surface of moon, the Kinetic energy of satellite equivalent to binding energy.
K.E. = B.E.
12mve2=GMmmR${\displaystyle \frac{1}{2}}m{v_e}^2={\displaystyle \frac{G{M}_{m}m}{R}}$
ve=2GMR−−−−−√$v_e=\sqrt{{\displaystyle \frac{2GM}{R}}}$..........(1)
We know that
gm=GMmR2$g_m={\displaystyle \frac{G{M}_m}{R^2}}$
hence,
ve=2gmR−−−−−√$v_e=\sqrt{2g_{m}R}$..........(2)

Result

Reason for presence of atmosphere on big planets

Big planets  have atmosphere as the velocity of molecules in its atmosphere [vrms=3RTM−−−−−√]$[v_{rms}=\sqrt{{\displaystyle \frac{3RT}{M}}}]$ is lesser than escape velocity [ve=2gR−−−−√]$[v_e=\sqrt{2gR}]$. This is why earth has atmosphere while moon has no atmosphere.

Example

Time Period of revolution of satellites

Example: Assume that a satellite is revolving around Earth in a circular orbit almost close to the surface of Earth. What is the time period of revolution of satellite is ($($Radius of earth is 6400 km, g=$g=$ 9.8 ms−2)$m{s}^{-2})$ ?

Solution: Time period of the satellite is given by T=2πrg−−√$T=2\pi \sqrt{{\displaystyle \frac{r}{g}}}$
Almost close to surface of Earth ⟹r=Re$⟹r=R_e$
∴$\therefore$ Time period T=2π6400×1039.8−−−−−−−−−√$T=2\pi \sqrt{{\displaystyle \frac{6400\times {10}^3}{9.8}}}$
Time period T=5077$T=5077$ seconds

Example

Time Period of satellites

Example: A geostationary satellite orbits around the earth in a circular orbit of radius 36000$36000$ km. Then, the time period of a spy satellite orbiting a few hundred kilometers above the earth's surface  will approximately be
($($ Given: REarth=6400$R_{Earth}=6400$ km )$)$

Solution:
For a satellite of mass m$m$ moving with a velocity v$v$ in a circular orbit of radius r$r$ around the earth of mass M$M$, we have
mv2r=GmMr2${\displaystyle \frac{m{v}^2}{r}}={\displaystyle \frac{GmM}{r^2}}$ or v=GMr−−−−√$v=\sqrt{{\displaystyle \frac{GM}{r}}}$
Now v=2πrT$v={\displaystyle \frac{2\pi r}{T}}$. Thus 2πrT=GMr−−−−√${\displaystyle \frac{2\pi r}{T}}=\sqrt{{\displaystyle \frac{GM}{r}}}$ or T∝r32$T\propto r^{{\displaystyle \frac{3}{2}}}$
∴T2T1=(r2r1)32$\therefore {\displaystyle \frac{T_2}{T_1}}=(\frac{r_2}{r_1}{)}^{{\displaystyle \frac{3}{2}}}$.....(1)
Given
r2=6400km$r_2=6400km$ and r1=36000km$r_1=36000km$. For a geostationary satellite
T1=24h$T_1=24h$. Using these values in (1), we have get T2=24×(64360)32=1.8h$T_2=24\times ({\displaystyle \frac{64}{360}}{)}^{{\displaystyle \frac{3}{2}}}=1.8h$.

Example

Orbital Speed of satellites

Example: The orbital speed for an Earth satellite near the surface of the Earth
is 7 km/sec. If the radius of the orbit is 4 times the radius of the Earth, what would be the orbital speed?

Solution: Orbital speed is v=GMer−−−−−√$v=\sqrt{{\displaystyle \frac{G{M}_e}{r}}}$ for a satellite at a distance r$r$ from the Earth's centre.
Firstly, the satellite is said to be near the earth's surface, we take the radius as Earth's radius, 'R$R$'.
⇒v=GMeR−−−−−√=7km/s$\Rightarrow v=\sqrt{{\displaystyle \frac{G{M}_e}{R}}}=7km/s$
Now, the radius of the orbit is 4 times the radius of the earth(R)$(R)$, the orbital speed (v′)$(v^{{}^{\prime }})$ is given by,
v′=GMe4R−−−−−√=14√GMeR−−−−−√=0.5×7=3.5$v^{{}^{\prime }}=\sqrt{{\displaystyle \frac{G{M}_e}{4R}}}={\displaystyle \frac{1}{\sqrt{4}}}\sqrt{{\displaystyle \frac{G{M}_e}{R}}}=0.5\times 7=3.5$ km/s

Example

Relative motion of geostationary satellite with respect to earth

Example: What is the relative velocity of geostationary satellite with respect to the spinning motion of the Earth?

Solution: Geostationary satellites remain at the same position with respect to the Earth, so that they scan the same place in a better way. Therefore, the relative velocity of geostationary satellite with respect to the spinning motion of the Earth is 0 m/s$0m/s$.

Example

Binary System of Stars

Example: Two small dense stars rotate about their common centre of mass as a binary system with the period 1$1$ year for each. One star is of double the mass of the other and the mass of the lighter one is 1/3$1/3$ of the mass of the sun. The distance between the stars is xR$xR$ if distance between the earth & the sun is R$R$, find x$x$.

Solution: Let, a$a$ be the distance between stars and their masses are 2M3,M3${\displaystyle \frac{2M}{3}},{\displaystyle \frac{M}{3}}$. M$M$ is mass of sun.
ωs=ωe=2π365×24×3600=(GMR3)0.5${\omega }_s={\omega }_e={\displaystyle \frac{2\pi }{365\times 24\times 3600}}=({\displaystyle \frac{GM}{R^3}}{)}^{0.5}$
consider the binary star system, Gravitational force = centripetal force
G(2M3)(M3)a2=M3(ωs)22a3${\displaystyle \frac{G({\displaystyle \frac{2M}{3}})({\displaystyle \frac{M}{3}})}{a^2}}={\displaystyle \frac{M}{3}}({\omega }_s{)}^2{\displaystyle \frac{2a}{3}}$
GMa2=GMaR3→a=R${\displaystyle \frac{GM}{a^2}}={\displaystyle \frac{GMa}{R^3}}\to a=R$
So,x=R$x=R$.

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