Physics · Gravity and Orbits

Escape Velocity

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Wrap-up

What you learned

Recommended next: Open concept testCheck whether the core ideas are ready without leaving this concept.
Read next: Circular Orbits and Orbital SpeedContrast escape with the speed that holds a bound circle.

Key takeaway

Escape velocity is the launch speed that makes total specific energy zero at the starting radius.
Gravity keeps slowing an escaping launch; escape means there is no finite turnaround radius, not that gravity vanishes.
Increasing source mass raises the threshold, while launching from a larger radius lowers it.
Being faster than circular speed is not enough for radial escape if the total specific energy is still negative.

Common misconception

Getting very far from the source is not the same as escaping. A high bound launch can travel far outward and still return if E/m remains negative.

Distance alone does not decide escape. The deciding quantity is the sign of $E / m$ . Zero or positive total specific energy means there is no finite turnaround radius, while negative total specific energy means the launch is still bound.

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Bound-orbit timingGravity and Orbits

Kepler's Third Law and Orbital Periods

Compare circular orbits around one source mass and see why larger orbits take longer: the path is longer, the circular speed is lower, and the same live model makes the period law visible without hiding the gravity-speed link.

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Circular Orbits and Orbital Speed

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Gravitational Potential and Potential Energy

See one source mass create a negative potential well, compare how potential and potential energy change with distance, and connect the downhill slope of phi to the gravitational field on the same live model.

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Equation snapshotSpecific total energy · Escape speedShow 3 equations

Read these as one test: the launch starts with kinetic energy, the well contributes negative potential energy, and the sign of E/m decides whether a finite turnaround exists.

Specific total energy
$df r a c E m = df r a c v^{2} 2 - df r a c M r$
The sign of this quantity decides the outcome: negative is bound, zero is threshold, and positive escapes.
Escape speed
$v_{t e x t esc} = s q r t df r a c 2 M r_{0}$
A deeper well needs a larger threshold speed, while a shallower starting point needs a smaller one.
Finite turnaround radius
$r_{t e x t t u r n} = df r a c M - (E / m) q q u a d (E < 0)$
When $E < 0$ , this gives the maximum radius before the bound launch must turn around.

Why it behaves this way

Explanation

Escape velocity is the minimum outward launch speed from a chosen radius that makes the total specific energy exactly zero. At that threshold, gravity still pulls inward and keeps slowing the launch, but there is no finite turnaround radius. Below that threshold, even a very long outward trip is still bound and must eventually return.

This lab keeps one source mass, one launch radius, one speed factor, and one live radial path on the same state. That lets you compare three ideas students often mix up: the local circular speed, the local escape speed, and the sign of the total specific energy. Going far from the source is not enough by itself. Escape is the case with no finite turnaround radius.

Key ideas

01Escape is decided by total specific energy, not by distance alone.

02With displayed units using

G = 1

, the escape speed from launch radius

r_{0}

v_{esc} = 2 M / r_{0}

. A heavier source raises the threshold, while a larger launch radius lowers it.

03The local circular speed is useful for comparison, but a launch can be faster than circular and still remain bound if its total specific energy stays negative.

Worked examples

Solve the live escape state

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Frozen walkthrough

Step through the frozen example

Frozen walkthrough

Use the current source mass, launch radius, and speed factor as evidence. First find the threshold launch speed. Then read the live kinetic, potential, and total specific energies to decide whether the launch must escape or eventually turn around.

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Example 1 of 2

Frozen valuesUsing frozen parameters

For the current source mass and launch radius, what outward launch speed makes the total specific energy exactly zero?

M

Source mass

4 kg

r_{0}

Launch radius

1.6 m

1. Set the total specific energy to zero

At escape threshold,

0 = df r a c v_{t e x t esc}^{2} 2 - df r a c M r_{0}

2. Solve for the threshold speed

v_{t e x t esc} = s q r t 2 M / r_{0} = s q r t 2 (4) /1.6

in the displayed

G = 1

units.

3. Compare the live escape and circular speeds

That gives

v_{t e x t esc} = 2.24, ma t h r m m / s

, while the circular-speed comparison is

v_{c} = 1.58, ma t h r m m / s

at the same radius.

Escape speed at launch

v_{t e x t esc} = 2.24, ma t h r m m / s

The escape threshold comes straight from setting the total energy to zero at the launch point.

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Gravity and Orbits

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