Orbital Velocity

Orbital Velocity

Vishal kumarUpdated on 02 Jul 2025, 07:03 PM IST

Orbital velocity refers to the speed at which an object must travel to remain in a stable orbit around a celestial body, such as Earth. This fundamental concept in physics and space science is crucial for understanding how satellites, space stations, and spacecraft navigate the cosmos. When an object achieves orbital velocity, it counteracts the pull of gravity by continuously falling toward the planet while moving forward, effectively balancing its trajectory. This principle is not only pivotal for launching satellites that provide crucial communication and weather data but also for ambitious space missions exploring our solar system. In everyday life, the technology and principles derived from orbital velocity enhance our global positioning systems, and satellite TV, and even contribute to scientific research that drives advancements in technology and exploration.

This Story also Contains

  1. What is Orbital Velocity?
  2. Proof
  3. Solved Examples Based on Orbital Velocity
  4. Summary

What is Orbital Velocity?

Orbital velocity is the speed at which an object must travel to remain in a stable orbit around a celestial body, such as a planet or star. For an object to achieve this velocity, it must balance the gravitational pull of the body it is orbiting with its own forward motion. Essentially, the object is in constant free-fall towards the celestial body but moving forward fast enough that the curvature of its path matches the curvature of the body’s surface.

For the figure below


$
\begin{aligned}
& r_1=r_p=a-c \\
& r_2=r_a=a+c
\end{aligned}
$
If Eccentricity is given by

$
\text { (e) }=\frac{c}{a}
$

then The velocity of the planet at Apogee and Perigee in terms of Eccentricity is given by

$
\begin{aligned}
V_a & =\sqrt{\frac{G M}{a}\left(\frac{1-e}{1+e}\right)} \\
V_p & =\sqrt{\frac{G M}{a}\left(\frac{1+e}{1-e}\right)} \\
V_A & =\text { The velocity of the planet at apogee } \\
V_p & =\text { Velocity of perigee }
\end{aligned}
$

Proof

Let the mass of the sun be M and the mass of the planet be m

Applying the law of conservation of angular momentum at perigee and apogee about the sun
$
\begin{aligned}
& m v_p r_p=m v_a r_a \\
\Rightarrow & \frac{v_p}{v_a}=\frac{r_a}{r_p}=\frac{a+c}{a-c}=\frac{1+e}{1-e}
\end{aligned}
$
Applying the conservation of mechanical energy at perigee and apogee

$
\begin{aligned}
& \frac{1}{2} m v_p^2-\frac{G M m}{r_p}=\frac{1}{2} m v_a^2-\frac{G M m}{r_a} \Rightarrow v_p^2-v_a^2=2 G M\left[\frac{1}{r_p}-\frac{1}{r_a}\right] \\
\Rightarrow \quad & v_a^2\left[\frac{r_a^2-r_2^2}{r_p^2}\right]^2=2 G M\left[\frac{r_a-r_p}{r_a r_p}\right] \quad\left[\text { As } v_p=\frac{v_a r_a}{r_p}\right]
\end{aligned}
$

Applying the conservation of mechanical energy at perigee and apogee

$
\begin{aligned}
& \frac{1}{2} m v_p^2-\frac{G M m}{r_p}=\frac{1}{2} m v_a^2-\frac{G M m}{r_a} \Rightarrow v_p^2-v_a^2=2 G M\left[\frac{1}{r_p}-\frac{1}{r_a}\right] \\
\Rightarrow & v_a^2\left[\frac{r_a^2-r_p^2}{r_p^2}\right]^2=2 G M\left[\frac{r_a-r_p}{r_a r_p}\right] \quad\left[\text { As } v_p=\frac{v_a r_a}{r_p}\right] \\
\Rightarrow & v_a^2=\frac{2 G M}{r_a+r_p}\left[\frac{r_p}{r_r}\right] \Rightarrow v_a^2=\frac{2 G M}{2 a}\left(\frac{a-c}{a+c}\right)=\frac{G M}{a}\left(\frac{1-e}{1+e}\right)
\end{aligned}
$
Thus the speeds of the planet at apogee and perigee are

$
v_a=\sqrt{\frac{G M}{a}\left(\frac{1-e}{1+e}\right)}, \quad v_p=\sqrt{\frac{G M}{a}\left(\frac{1+e}{1-e}\right)}
$

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Solved Examples Based on Orbital Velocity

Example 1: The ratio of velocities of a planet at perigee and apogee $\frac{v_p}{v_a}$ is equal to

1) 1
2) $\frac{1-e}{1+e}$
3) $\frac{1+e}{1-e}$
4) $\frac{1+e^2}{1-e^2}$

Solution:

The velocity of the planet in terms of Eccentricity

$\begin{aligned} & V_a=\sqrt{\frac{G M}{a}\left(\frac{1-e}{1+e}\right)} \\ & V_p=\sqrt{\frac{G M}{a}\left(\frac{1+e}{1-e}\right)} \\ & V_A=\text { The velocity of the planet at apogee } \\ & V_p=\text { Velocityo fperigee } \\ & \text { wherein } \\ & \text { Eccentricity (e) }=\frac{c}{a} \\ & r_p=a-c \\ & r_a=a+c\end{aligned}$

By conservation of angular momentum,

$\begin{aligned} & m v_p r_p=m v_a r_a \\ & \Rightarrow \frac{v_p}{v_a}=\frac{r_a}{r_p}=\frac{a+c}{a-c}=\frac{1+e}{1-e}\end{aligned}$

Hence, the answer is the option (3).

Example 2: A satellite is in an all-optical orbit around planet P. It is observed that the velocity of the satellite when it is farthest from the planet is 6 times less than that when it is closest to the planet. The ratio of distances b/w the satellite and the planet at closest and farthest point is

1) $3:4$

2) $1:6$

3) $1:2$

4) $1:3$

Solution:

As the angular momentum remains conserved. So,
$
\begin{aligned}
& L_{\text {Nearest }}=L_{\text {Farthest }} \\
& m \times v_n \times r_n=m \times v_f \times r_f
\end{aligned}
$
Also,

$
V_f=\frac{V_n}{6}
$
As,

$
\begin{aligned}
& V_n \times r_n=V_f \times r_f \\
& V_n \times r_n=\frac{V_n}{6} \times r_f
\end{aligned}
$
So,

$
\frac{r_n}{r_f}=\frac{1}{6}
$

Hence, the answer is the option (2).


Summary

Orbital velocity is the speed required for an object to maintain a stable orbit around a celestial body, balancing gravitational pull with forward motion. It is determined using the formula v = $\sqrt{\frac{G M}{r}}$. where G is the gravitational constant, M is the mass of the celestial body, and r is the distance from its centre. For elliptical orbits, the velocities at apogee (farthest point) and perigee (closest point) are given by $V_a=\sqrt{\frac{G M}{a}\left(\frac{1-e}{1+e}\right)}$ and $V_p=\sqrt{\frac{G M}{a}\left(\frac{1+e}{1-e}\right)}$ respectively. Examples illustrate how these velocities relate to orbital parameters and conservation laws.

Frequently Asked Questions (FAQs)

Q: What is a "hyperbolic trajectory" and how does its velocity compare to orbital velocity?
A:
A hyperbolic trajectory is an open orbit where the object's velocity exceeds escape velocity. Its speed is always greater than the orbital velocity for a circular orbit at the same distance, allowing the object to escape the gravitational influence of the central body.
Q: How does the concept of "escape velocity" differ from orbital velocity?
A:
Escape velocity is the minimum speed needed to break free from a body's gravitational influence, while orbital velocity is the speed needed to maintain a stable orbit. Escape velocity is always √2 times the circular orbital velocity at any given altitude.
Q: What is the concept of "orbital period" and how is it related to orbital velocity?
A:
Orbital period is the time taken for an object to complete one orbit. It's inversely related to orbital velocity: higher velocity results in shorter periods. For circular orbits, the period T is given by T = 2π√(r³/GM), where r is the orbital radius.
Q: How does Earth's oblateness (non-spherical shape) affect satellite orbital velocities?
A:
Earth's oblateness causes variations in gravitational force at different latitudes, which can affect orbital velocities. This leads to orbital precession and can be used advantageously in sun-synchronous orbits or must be compensated for in other types of orbits.
Q: How does the concept of "delta-v" relate to changes in orbital velocity?
A:
Delta-v represents the change in velocity needed to perform an orbital maneuver. It's a crucial concept in space mission planning, as it directly relates to the amount of fuel needed for orbital transfers, inclination changes, and interplanetary trajectories.
Q: How does the concept of "specific orbital energy" relate to orbital velocity?
A:
Specific orbital energy is the sum of kinetic and potential energy per unit mass of an orbiting object. It's directly related to orbital velocity: as specific orbital energy increases, so does the object's ability to achieve higher orbits or escape velocity.
Q: How do space agencies use "gravity assists" to change a spacecraft's orbital velocity?
A:
Space agencies use gravity assists to change a spacecraft's velocity and trajectory by having it fly close to a planet. The spacecraft's path bends and its speed changes relative to the Sun, allowing it to gain or lose orbital energy without using fuel.
Q: How does the concept of orbital velocity apply to binary star systems?
A:
In binary star systems, both stars orbit their common center of mass. Their orbital velocities depend on their masses and the distance between them, following the same principles as planets orbiting the Sun or satellites orbiting Earth.
Q: What is the concept of "orbital inclination" and how does it affect orbital velocity?
A:
Orbital inclination is the angle between the orbital plane and the equatorial plane of the central body. While inclination doesn't directly affect orbital velocity, changing inclination requires significant velocity changes, making it one of the most energy-intensive orbital maneuvers.
Q: How does the "Oberth effect" allow for more efficient changes in orbital velocity?
A:
The Oberth effect allows for more efficient velocity changes when a spacecraft is at its highest speed in an orbit (usually at perigee). Burning fuel at this point provides a greater change in kinetic energy and, therefore, a larger change in velocity for the same amount of fuel.