Orbital Velocity of Satellite, Time Period, Height of Satellite and Verification of Kepler's Third Law
Orbital Velocity of Satellite
The minimum velocity required by a satellite to remain in a stable circular orbit around the Earth is called orbital velocity.
Derivation of Orbital Velocity
Consider:
- Mass of Earth = \(M\)
- Mass of Satellite = \(m\)
- Orbital Radius = \(r\)
- Orbital Velocity = \(v_0\)
Gravitational Force on Satellite:
$$ F_g=\frac{GMm}{r^2} $$Centripetal Force Required:
$$ F_c=\frac{mv_0^2}{r} $$For stable circular motion,
$$ F_g=F_c $$ $$ \frac{GMm}{r^2}=\frac{mv_0^2}{r} $$Cancelling \(m\),
$$ \frac{GM}{r^2}=\frac{v_0^2}{r} $$ $$ v_0^2=\frac{GM}{r} $$ $$ v_0=\sqrt{\frac{GM}{r}} $$For a satellite,
$$ r=R+h $$where \(R\) is the radius of Earth and \(h\) is the height of the satellite above Earth's surface.
Therefore,
$$ v_0=\sqrt{\frac{GM}{R+h}} $$Using
$$ g=\frac{GM}{R^2} $$or
$$ GM=gR^2 $$we get
$$ v_0=\sqrt{\frac{gR^2}{R+h}} $$Satellite Close to Earth's Surface
If \(h=0\), then
$$ v_0=\sqrt{gR} $$Substituting
$$ g=9.8\;m\,s^{-2} $$ $$ R=6.4\times10^6\;m $$ $$ v_0=7.92\times10^3\;m\,s^{-1} $$ $$ v_0=7.92\;km\,s^{-1} $$Time Period of a Satellite
The time taken by a satellite to complete one revolution around the Earth is called its time period.
Distance travelled in one revolution:
$$ 2\pi r $$Therefore,
$$ T=\frac{2\pi r}{v_0} $$Substituting
$$ v_0=\sqrt{\frac{GM}{r}} $$ $$ T=\frac{2\pi r}{\sqrt{\frac{GM}{r}}} $$ $$ T=2\pi\sqrt{\frac{r^3}{GM}} $$Time Period Formula:
$$ T=2\pi\sqrt{\frac{r^3}{GM}} $$Verification of Kepler's Third Law
From the time period equation,
$$ T=2\pi\sqrt{\frac{r^3}{GM}} $$Squaring both sides,
$$ T^2=\frac{4\pi^2r^3}{GM} $$Rearranging,
$$ \frac{T^2}{r^3}=\frac{4\pi^2}{GM} $$Since \(\frac{4\pi^2}{GM}\) is constant for Earth,
$$ T^2\propto r^3 $$Hence, the square of the time period of a satellite is directly proportional to the cube of the radius of its orbit.
Therefore, Kepler's Third Law is verified.
Height of a Satellite Above Earth's Surface
From
$$ T=2\pi\sqrt{\frac{r^3}{GM}} $$Squaring,
$$ T^2=\frac{4\pi^2r^3}{GM} $$ $$ r^3=\frac{GMT^2}{4\pi^2} $$Taking cube root,
$$ r=\left(\frac{GMT^2}{4\pi^2}\right)^{1/3} $$Since
$$ r=R+h $$Therefore,
$$ R+h=\left(\frac{GMT^2}{4\pi^2}\right)^{1/3} $$Hence,
$$ h=\left(\frac{GMT^2}{4\pi^2}\right)^{1/3}-R $$Relation Between Escape Velocity and Orbital Velocity
Escape Velocity:
$$ v_e=\sqrt{2gR} $$Orbital Velocity:
$$ v_0=\sqrt{gR} $$Dividing,
$$ \frac{v_e}{v_0} = \frac{\sqrt{2gR}}{\sqrt{gR}} $$ $$ \frac{v_e}{v_0} = \sqrt{2} $$Therefore,
$$ v_e=\sqrt{2}\,v_0 $$Thus, escape velocity is \(\sqrt{2}\) times the orbital velocity.
Frequently Asked Questions (FAQs)
1. What is orbital velocity?
Orbital velocity is the minimum velocity required by a satellite to remain in a stable orbit around the Earth.
2. What is the orbital velocity near Earth's surface?
$$ v_0=7.92\;km\,s^{-1} $$3. Does orbital velocity depend on the mass of the satellite?
No, orbital velocity is independent of the mass of the satellite.
4. State Kepler's Third Law.
$$ T^2\propto r^3 $$5. What is the relation between escape velocity and orbital velocity?
$$ v_e=\sqrt{2}\,v_0 $$Multiple Choice Questions (MCQs)
1. The orbital velocity of a satellite is
A. \(\sqrt{\frac{r}{GM}}\)
B. \(\sqrt{\frac{GM}{r}}\)
C. \(\frac{GM}{r}\)
D. \(\sqrt{GMr}\)
Answer: B
2. Orbital velocity near Earth's surface is
A. 11.2 km/s
B. 9.8 km/s
C. 7.92 km/s
D. 5 km/s
Answer: C
3. Kepler's Third Law is
A. \(T\propto r\)
B. \(T\propto r^2\)
C. \(T^2\propto r^3\)
D. \(T^3\propto r^2\)
Answer: C
4. Escape velocity is _____ times orbital velocity.
A. 1
B. \(\sqrt{2}\)
C. 2
D. 4
Answer: B
5. The orbital radius of a satellite is
A. \(R-h\)
B. \(R+h\)
C. \(Rh\)
D. \(\frac{R}{h}\)
Answer: B
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