HSC Physics

Module 5: Advanced Mechanics5.1 Projectile Motion

5.2 Circular Motion

5.3 Motion in Gravitational Fields2 Topics

Module 6: Electromagnetism6.1 Charged Particles, Conductors and Electric and Magnetic Fields

6.2 The Motor Effect1 Topic

6.3 Electromagnetic Induction

6.4 Applications of the Motor Effect1 Topic

Module 7: The Nature of Light7.1 Electromagnetic Spectrum3 Topics

7.2 Light: Wave Model

7.3 Light: Quantum Model2 Topics

7.4 Light and Special Relativity

Module 8: From the Universe to the Atom8.1 Origins of the Elements5 Topics

8.2 Structure of the Atom3 Topics

8.3 Quantum Mechanical Nature of the Atom2 Topics

8.4 Properties of the Nucleus2 Topics

8.5 Deep Inside the Atom4 Topics
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 EduKits Education
 Michael
Escape velocity is the minimum velocity required to escape the gravitational field of a body from its surface.
v_{\text{esc}} = \sqrt{\frac{2GM}{r}}
An object has ‘escaped’ from a gravitational field when its total orbital energy is equal to zero. This is because the object must have enough kinetic energy to continue moving away until its gravitational potential energy is equal to zero. Otherwise, the object will return from whence it came.
Begin the derivation by equating the total orbital energy to zero.
K+U=0
Next, substitute the relevant formulae for both the kinetic (K) and potential (U) elements of the orbital energy, as seen below. The large M refers to the mass of the larger body, while the smaller m refers to the escaping object.
\frac{1}{2}mv_{\text{esc}}^2  \frac{GMm}{r}=0
Finally, cancel m and rearrange the equation. Take the square root of both sides to arrive at the final equation for escape velocity.
v_{\text{esc}}^2 = \frac{2GM}{r} \space \space \rightarrow \space \space v_{\text{esc}} = \sqrt{\frac{2GM}{r}}