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Animated Solution for Physics - Gravitation: A body is moving in a low circular orbit about a planet of mass $M$ and radius $R$ . The radius of the orbit can be taken to be $R$ itself. Then, the ratio of the speed of this body in the orbit to the escape velocity from the planet is

Q: Is escape velocity always $\sqrt{2}$ times orbital velocity?

Yes, for a given distance $r$ from the center of the massive body, the escape velocity from that point is always exactly $\sqrt{2}$ times the circular orbital velocity at that same distance.

Q: What happens if the satellite's speed is increased but not up to the escape velocity?

If the speed is increased beyond $v_o$ but remains less than $v_e$, the circular orbit will stretch into an elliptical orbit. If it reaches or exceeds $v_e$, the path becomes parabolic or hyperbolic, respectively, and it escapes.

Select Answer:

Visualized Solution

Visualizing the Planet

Let the mass of the planet be $M$ .
Let the radius of the planet be $R$ .

The Low Circular Orbit

A body of mass $m$ is in a low circular orbit.
Orbital radius, $r \approx R$ .

Orbital Velocity ( $v_{o}$ )

The orbital velocity $v_{o}$ of a satellite revolving close to the surface is given by:
$v_{o} = \frac{GM}{R}$

Escape Velocity ( $v_{e}$ )

The escape velocity $v_{e}$ from the surface of the planet is given by:
$v_{e} = \frac{2 GM}{R}$

Calculating the Ratio

We need to find the ratio $\frac{v _{o}}{v _{e}}$ .
$\frac{v _{o}}{v _{e}} = \frac{\frac{GM}{R}}{\frac{2 GM}{R}}$

Final Answer

$\frac{v _{o}}{v _{e}} = \frac{1}{2}$

The Way Forward

What if the satellite was at a height $h = R$ ?
How would the ratio $\frac{v _{o}}{v _{e}}$ change?

00:00 / 00:00

The Sigma Insight: Orbital Motion of a Satellite

Solution Diagram

The Dance of Gravity

Orbiting vs Escaping

Imagine you are standing on a massive planet of mass

M

and radius

R

. You throw a stone horizontally. It falls back. You throw it harder, it falls further away. If you throw it just hard enough, the curvature of its fall will perfectly match the curvature of the planet, and it will keep falling forever without ever hitting the ground. This is the essence of an orbit!

For a satellite in a low circular orbit, it is skimming just above the atmosphere. The distance from the center of the planet to the satellite is practically just the radius of the planet, so

r \approx R

The Orbital Velocity ( $v_{o}$ )

To maintain this circular dance, the gravitational pull of the planet must provide the exact centripetal force required.

Equating the two forces:

\frac{GM m}{R ^{2}} = \frac{m v _{o}^{2}}{R}

Solving for the orbital velocity

v_{o}

, we get:

v_{o} = \frac{GM}{R}

This is the speed required to just stay in orbit. But what if the satellite wants to break free entirely?

The Escape Velocity ( $v_{e}$ )

To escape the gravitational clutches of the planet forever, the satellite needs enough kinetic energy to overcome the gravitational potential energy binding it to the planet. It needs to reach infinity with at least zero speed.

By conservation of energy:

\frac{1}{2} m v_{e}^{2} - \frac{GM m}{R} = 0

Solving for the escape velocity

v_{e}

, we get:

v_{e} = \frac{2 GM}{R}

Notice something beautiful here? The escape velocity is exactly

2

times the orbital velocity!

The Final Ratio

The question asks for the ratio of the orbital speed to the escape velocity. Let's divide them:

\frac{v _{o}}{v _{e}} = \frac{\frac{GM}{R}}{\frac{2 GM}{R}}

The

\frac{GM}{R}

term cancels out beautifully, leaving us with:

\frac{v _{o}}{v _{e}} = \frac{1}{2}

Physical Insight: This means that if a satellite is in a low Earth orbit, you don't need to give it an infinite amount of energy to escape. You just need to increase its speed by a factor of

2

(about

41.4%

increase). This is a profound and elegant result of Newtonian mechanics!

Common Traps & FAQs

Why did we take the orbital radius as

R

instead of

R + h

The problem explicitly states it's a 'low circular orbit' and that 'the radius of the orbit can be taken to be

R

itself'. This means the height

h

above the surface is negligible compared to the planet's radius

R

(

h ≪ R

). Thus,

r = R + h \approx R

Does the mass of the satellite affect these velocities?

No, both orbital and escape velocities are independent of the mass

m

of the orbiting or escaping body. The mass

m

cancels out when equating forces or energies.

What does the ratio

1/ 2

signify physically?

It means that to escape the planet's gravity from a low orbit, a satellite needs its kinetic energy to be doubled (since

K \propto v^{2}

), which means its speed must increase by a factor of

2

Is escape velocity always

2

times orbital velocity?

Yes, for a given distance

r

from the center of the massive body, the escape velocity from that point is always exactly

2

times the circular orbital velocity at that same distance.

What happens if the satellite's speed is increased but not up to the escape velocity?

If the speed is increased beyond

v_{o}

but remains less than

v_{e}

, the circular orbit will stretch into an elliptical orbit. If it reaches or exceeds

v_{e}

, the path becomes parabolic or hyperbolic, respectively, and it escapes.

Animated Solution for Physics - Gravitation: A body is moving in a low circular orbit about a planet of mass $M$ and radius $R$ . The radius of the orbit can be taken to be $R$ itself. Then, the ratio of the speed of this body in the orbit to the escape velocity from the planet is

Select Answer:

Visualized Solution

Visualizing the Planet

The Low Circular Orbit

Orbital Velocity ( $v_{o}$ )

Escape Velocity ( $v_{e}$ )

Calculating the Ratio

Final Answer

The Way Forward

The Sigma Insight: Orbital Motion of a Satellite

The Dance of Gravity

The Orbital Velocity ( $v_{o}$ )

The Escape Velocity ( $v_{e}$ )

The Final Ratio

Similar Questions

A satellite of mass $m$ revolves around the earth of radius $R$ at a height $x$ from its surface. If $g$ is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is

List-I

List-II

Two satellites $A$ and $B$ have masses $m$ and $2 m$ respectively. $A$ is in a circular orbit of radius $R$ and $B$ is in a circular orbit of radius $2 R$ around the earth. The ratio of their kinetic energies, $T_{A} / T_{B}$ is

A satellite is in an elliptical orbit around a 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 between the satellite and the planet at closest and farthest points is

A test particle is moving in a circular orbit in the gravitational field produced by mass density $ρ (r) = \frac{K}{r ^{2}}$ . Identify the correct relation between the radius $R$ of the particle's orbit and its period $T$

Select Answer:

Visualized Solution

Visualizing the Planet

The Low Circular Orbit

Orbital Velocity ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vo​)

Escape Velocity ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ve​)

Calculating the Ratio

Final Answer

The Way Forward

The Sigma Insight: Orbital Motion of a Satellite

The Dance of Gravity

The Orbital Velocity ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vo​)

The Escape Velocity ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ve​)

The Final Ratio

Similar Questions

List-I

List-II

Orbital Velocity ( $v_{o}$ )

Escape Velocity ( $v_{e}$ )

The Orbital Velocity ( $v_{o}$ )

The Escape Velocity ( $v_{e}$ )