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Escape Velocity
Escape velocity is the minimum initial speed required for an object to escape permanently from a gravitational field without further propulsion, reaching infinity with zero speed in the ideal model.
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Escape Velocity
What is it?
Escape velocity is the minimum initial speed required for an object to escape permanently from a gravitational field without further propulsion, reaching infinity with zero speed in the ideal model.
In simple words
- Mass of escaping object cancels.
Recommended for Class 12
Full school-level explanation
Escape velocity is the minimum initial speed required for an object to escape permanently from a gravitational field without further propulsion, reaching infinity with zero speed in the ideal model.
Important Result
Consider object mass `m` launched from radius `R`. Initial total energy: Ei = ½mve² − GMm/R At minimum escape condition, at infinity: `U=0` `K=0` so: Ef=0 By energy conservation: ½mve² − GMm/R = 0 Therefore: ½mve² = GMm/R Cancel `m`: ve² = 2GM/R Thus: ve = √(2GM/R) Since surface orbital speed is: vo = √(GM/R) therefore: ve = √2 vo Escape velocity is independent of the escaping object’s mass. For Earth’s surface, ignoring atmosphere and Earth rotation: approximately: 11.2 km/s
Formula / Key Relation
**ve = √(2GM/R)**
Remember
- 11.2 km/s
- ve = √(2GM/R)
Deeper understanding
Go deeper on Escape Velocity: connect the definition to reasons, relationships, exceptions, and exam wording.
Common mistakes
Exam focus
ESCAPE
Exam focus
Quick revision
- 11.2 km/s
- ve = √(2GM/R)
- Important Result: Consider object mass `m` launched from radius `R`. Initial total energy: Ei = ½mve² − GMm/R
- Formula / Key Relation: **ve = √(2GM/R)**
- Mass of escaping object cancels.
For long answers on Escape Velocity: define clearly → explain the main parts → give one example → state one exam caution.
Exam Focus
Exam Focus
Important points
- 11.2 km/s
- ve = √(2GM/R)
- Important Result: Consider object mass `m` launched from radius `R`. Initial total energy: Ei = ½mve² − GMm/R
- Formula / Key Relation: **ve = √(2GM/R)**
Important formulas
- 11.2 km/s
- ve = √(2GM/R)
Common exam traps
- Escape velocity does not mean gravity becomes zero.
- The object continues to experience gravity while moving outward.
Long question structure
Definition → key parts → example → exam caution
Common Mistakes
Common Mistakes
- Mistake
Escape velocity does not mean gravity becomes zero.
- Mistake
The object continues to experience gravity while moving outward.
- Mistake
Escape velocity does not mean gravity becomes zero.
Revision Zone
60-Second Revision
Quick cues from this topic — use them for last-minute revision.
11.2 km/s
ve = √(2GM/R)
Important Result: Consider object mass `m` launched from radius `R`. Initial total energy: Ei = ½mve² − GMm/R
Formula / Key Relation: **ve = √(2GM/R)**
**Escape velocity is the minimum initial speed required for an object to escape permanently from a gravitational field without further propulsion, reaching infi
ve = √(2GM/R)
- Definition
- **Escape velocity is the minimum initial speed required for an object to escape permanently from a gravitational field without further propulsion, reaching infi
- Formula
- ve = √(2GM/R)
- Critical warning
- Escape velocity does not mean gravity becomes zero.
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Exam Practice
Define
Q1. Define Escape Velocity.
Explain
Q2. Explain: Important Result.
Example
Q3. Give one clear example of Escape Velocity and say why it fits.
Mistake
Q4. State one common mistake students make about Escape Velocity.
Exam note
Q5. Write a short exam note on Escape Velocity: definition, two points, one example.
Sources & References
School-standard facts for Matric and Intermediate. Original explanations — not copied from textbooks.