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Orbital velocity formula is the velocity that helps a natural or artificial satellite to remain in its orbit. Orbital Velocity formula is referred to as the formula required to calculate the velocity of a body that revolves around another body. Orbital velocity is applied in the artificial satellites as it helps them revolve around a specific planet. Orbital velocity of any planet can be calculated with the known mass M and radius R. It is measured in meters per second (m/s).
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Key Terms: Gravitational Constant, Orbital Velocity, Centripetal Force, Satellite, Orbit, Velocity
What is Orbital Velocity?
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Orbital velocity of a body can be defined as the velocity at which it is seen to revolve around another body. Objects that revolve around the Earth in a uniform circular motion are referred to be in an Orbit.
Orbital Velocity
The orbit’s velocity depends upon the distance between the object and the earth’s centre. Orbital velocity of any planet can be calculated with the known mass M and radius R. It is measured in meters per second (m/s).
Read More: Differences Between Acceleration and Velocity
Orbital Velocity Formula
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Orbital velocity formula is given as:
| Vorbit = \(\sqrt{\frac{GM}{R}}\) |
Where,
- G = Gravitational Constant
- M = Mass of the Body at Centre
- R = Radius of the Orbit
Frequently Asked Questions on Orbital VelocityQues. Is Orbital Velocity affected by mass? (1 mark) Ans. Considering the mass of a satellite (not the mass of the body which orbits), then the mass is seen to not get affected by orbital velocity. Ques. What factors affect Orbital Velocity? (3 marks) Ans. The factors that affect Orbital Velocity include:
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Derivation of Orbital Velocity Formula
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Orbital Velocity formula can be calculated by first knowing the gravitational force and centripetal force. While centripetal force is the force that is required to possess the circular motion, the gravitational force is exhibited by a central body on the other body that is in orbit so that the orbiting body remains in its orbit.
- To obtain the Orbital Velocity Formula, let’s consider a satellite with the mass m revolving in a circular orbit around the Earth with a radius r and at a height h from the Earth’s surface.
- Now, let R and M be the radius and mass of the earth respectively.
Then, it can be said:
| ⇒ R = R+h |
- A centripetal force m\(v_0^2\)r is required to help revolve the satellite by the gravitational force \(G\frac{Mm}{r^2}\). This force is applied between the Earth and the satellite.
- Now, after equating the centripetal force and gravitational force, we have:
⇒ \(\frac{mv_0^2}{r}\) = \(G\frac{Mm}{r^2}\)
⇒ \(v_0^2\) =\(\frac{GM}{r}\) = \(\frac{GM}{R+h}\) . . . (1)
Because the acceleration of the gravity(g) is also acting, we have:
⇒ GM = gR2
Now,
⇒ v0 = \(\sqrt{\frac{gR^2}{R+h}}\)
Simplifying the equation, we have:
⇒ v0 = RgR + h
Let us have h as the height from the surface and the acceleration due to gravity as g’. Then,
⇒ g’ = \(\frac{GM}{(R +h)^2}\)
⇒ \(\frac{GM}{(R +h)}\) = g’ (R+h) = g’r … (2)
Inserting the value of equation 2 in equation 1, we get:
| ⇒ v0 = \(\sqrt{g'r}\) = \(\sqrt{g'(R+h)}\) |
Previous Year Questions
- A satellite is revolving in a circular orbit at a height 'h' from … [JEE Mains 2016]
- A small satellite is revolving near earths surface …
- A satellite is launched in a circular orbit of radius R … [KEAM]
- What is a period of revolution of earth satellite … [KCET 2014]
- A satellite in a circular orbit of radius R has a period of 4 hours … [KCET 1996]
- For a satellite moving in an orbit around the earth … [NEET 2005]
- A satellite in force free space sweeps stationary interplanetary dust … [NEET 1994]
- A geostationary satellite is orbiting the earth at a height … [NEET 2012]
- A satellite of mass m is orbiting the earth (of radius R) at a height h … [NEET 2016]
- A remote-sensing satellite of earth revolves in a circular orbit … [NEET 2015]
- For a satellite escape velocity is 11 km/s … [NEET 1989]
- A satellite A of mass m is at a distance of r from the surface of the earth … [NEET 1993]
- A synchronous relay satellite reflects TV signals … [BHU UET]
- The time period of a satellite of earth is 5 hours … [BITSAT 2019]
- A artificial satellite is going round the earth assumed to be uniform sphere … [JIPMER]
- The orbit of geo-stationary satellite is circular, the time period of satellite … [BITSAT 2008]
- Magnitude of binding energy of satellite is … [UPSEE 2016]
Also Read: NCERT Solutions for Class 6 to 12 PDFs
Things to Remember
- To calculate the velocity of anybody that revolves around the other body, we require the orbital velocity.
- The formula to calculate the orbital velocity is Vorbit = \(\sqrt{\frac{GM}{R}}\) .
- To derive the formula of orbital velocity, the two things required are the gravitational force and centripetal force.
- The formula of centripetal force is \(\frac{mv_0^2}{r}\).
- The formula of gravitational force is G\(\frac{Mm}{r^2}\).
- The value of Gravitational constant is 6.674 x 10-11.
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Sample Questions
Ques. If a satellite revolves around the earth with a radius of earth R as 6.5x106m, the gravitational constant G as 6.67408 × 10-11 m3 kg-1 s-2 and the mass of the earth M as 5.9722×1024 kg. Estimate the orbital velocity of the earth. (3 marks)
Ans. Given:
R = 6.5 × 106 m
M = 5.9722×1024 kg
G = 6.67408 × 10-11 m3 kg-1 s-2
The Orbital velocity formula is given by,
Velocity of orbit = \(\sqrt{\frac{GM}{R}}\)
Ques. To study the planet Jupiter, a satellite has been launched. Estimate the velocity of the satellite as it orbits around the Jupiter with radius of the Jupiter as R = 70.5 × 106 m, the gravitational constant as G = 6.67408 × 10-11 m3 kg-1 s-2 and the Mass of Jupiter M as 1.5 × 1027 Kg. (2 marks)
Ans. As we already know,
Velocity of the Orbit= \(\sqrt{\frac{GM}{R}}\)
Ques. Estimate the period of revolution of the moon that revolves around the earth when the radius of the earth is given as 6400 km, the distance of the moon from the is around 3.84x105 km and the value of g as 9.8 m/s2. (4 marks)
Ans. To find:
Period of Revolution (T) = ?
Given:
R = radius of Earth = 6400 km = 6.4x108 m
r = radius of orbit of moon =3.84x105 km = 3.84x108 m
g = 9.8 m/s2
Velocity of the Orbit= \(\sqrt{\frac{GM}{R}}\)
Therefore, the period of revolution is 27.3 days.
Ques. Estimate the linear velocity of the moon if it takes 27 days to complete a revolution around the Earth and the distance between the two is given as 3.8x105 km. (3 marks)
Ans. To find:
vc= critical/linear velocity=?
Given:
T= Period of Revolution = 27 days = 27 x 24 x 60 x 60s
r = radius of orbit of moon =3.8 x 105 km = 3.8 x 108 m
Velocity of the Orbit= \(\sqrt{\frac{GM}{R}}\)
= 1.023 km/s
Therefore, the linear velocity of the moon is estimated around 1.023 km/s.
Ques. The estimated day for Venus to orbit around the sun is 225 days. Estimate the orbital radius and the speed of Venus when the mass of the sun is given as 2x1030 kg and the G= 6.67x10-11 in S.I. units. (5 marks)
Ans. To find:
r = Radius of orbit = ?
vc = Orbital Velocity = ?
Given:
M= Mass of the Sun= 2 x 1030 kg
T= Period of Venus = 225 days= 225 x 24 x 60 x 60s
G= 6.67x10-11 S.I. units.
Velocity of the Orbit= \(\sqrt{\frac{GM}{R}}\)
T = \(2 \Pi \sqrt{\frac{r^3}{GM}}\)
r = 3.841 x 108 m
r = 3.841 x 105 x 103 m
r = 3.841 x 105 km
Therefore, the orbital speed of Venus is 3.506x104 m/s and the Orbital radius of Venus is 1.085 x 1011 m.
Ques. State the formula of centripetal force and the gravitational force. (1 mark)
Ans. The formula of centripetal force is \(\frac{mv_0^2}{r}\) while the formula of gravitational force is G\(\frac{Mm}{r^2}\)
Ques. What is Orbital Velocity? State with the formula. (2 marks)
Ans. Orbital Velocity is the velocity that is required by a body to revolve around the other body. It can be seen in the artificial satellite that revolves around the planets. The formula of orbital velocity is Vorbit = \(\sqrt{\frac{GM}{R}}\) .
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