Revision NotesExam QuestionsFlashcards
IBPhysicsHLRevision NotesFieldsGravitational FieldsEscape Speed (HL)

Syllabus Edition

First teaching 2023

First exams 2025

|

Escape Speed (HL) (HL IB Physics)

Revision Note

Katie M

Author

Katie M

Last updated

Escape Speed

  • To escape a gravitational field, a mass must travel at, or above, the minimum escape speed
    • This is dependent on the mass and radius of the object creating the gravitational field, such as a planet, a moon or a black hole
  • Escape speed is defined as:

The minimum speed that will allow an object to escape a gravitational field with no further energy input

  • It is the same for all masses in the same gravitational field
    • For example, the escape speed of a rocket is the same as a tennis ball on Earth
  • The escape speed of an object is the speed at which all its kinetic energy has been transferred to gravitational potential energy
  • This is calculated by equating the equations:

1 half m v subscript e s c end subscript squared space equals space fraction numerator G M m over denominator r end fraction

  • Where:
    • m = mass of the object in the gravitational field (kg)
    • v subscript e s c end subscript = escape velocity of the object (m s−1)
    • G = Newton's Gravitational Constant
    • M = mass of the object to be escaped from (i.e. a planet) (kg)
    • r = distance from the centre of mass M (m)
  • Since mass m is the same on both sides of the equation, it can cancel on both sides of the equation:

1 half v subscript e s c end subscript squared space equals space fraction numerator G M over denominator r end fraction

  • Multiplying both sides by 2 and taking the square root gives the equation for escape velocity v subscript e s c end subscript:

v subscript e s c end subscript space equals space square root of fraction numerator 2 G M over denominator r end fraction end root

Escape Velocity Diagram, downloadable AS & A Level Physics revision notes

For an object to leave the Earth's gravitational field, it will have to travel at a speed greater than the Earth's escape velocity, v

  • Rockets launched from the Earth's surface do not need to achieve escape velocity to reach their orbit around the Earth
  • This is because:
    • They are continuously given energy through fuel and thrust to help them move
    • Less energy is needed to achieve orbit than to escape from Earth's gravitational field
  • The escape velocity is not the velocity needed to escape the planet but to escape the planet's gravitational field altogether
    • This could be quite a large distance away from the planet

Worked example

Calculate the escape speed at the surface of the Moon.

  • Density of the Moon = 3340 kg m−3
  • Mass of the Moon = 7.35 × 1022 kg

Answer:

Step 1: List the known quantities

  • Gravitational constant, G = 6.67 × 10−11 N m2 kg−2
  • Density of the Moon, ρ = 3340 kg m−3
  • Mass of the Moon, M = 7.35 × 1022 kg

Step 2: Rearrange the density equation for radius r

Density:  space rho space equals space M over V  and volume of a sphere:  V space equals space 4 over 3 straight pi r cubed

space rho space equals space fraction numerator M over denominator fraction numerator 4 over denominator 3 end fraction straight pi r cubed end fraction space equals space fraction numerator 3 M over denominator 4 straight pi r cubed end fraction

r space equals space cube root of fraction numerator 3 M over denominator 4 straight pi rho end fraction end root

Step 3: Calculate the radius by substituting in the values

r space equals space cube root of fraction numerator 3 space cross times space left parenthesis 7.35 cross times 10 to the power of 22 right parenthesis over denominator 4 straight pi space cross times space 3340 end fraction end root space equals space 1.7384 space cross times space 10 to the power of 6 space straight m

Step 4: Substitute r into the escape speed equation

v subscript e s c end subscript space equals space square root of fraction numerator 2 G M over denominator r end fraction end root space equals space square root of fraction numerator 2 space cross times space left parenthesis 6.67 cross times 10 to the power of negative 11 end exponent right parenthesis space cross times space left parenthesis 7.35 cross times 10 to the power of 22 right parenthesis over denominator 1.7384 cross times 10 to the power of 6 end fraction end root

Escape speed of the Moon:  v subscript e s c end subscript = 2.37 km s−1

Examiner Tip

When writing the definition of escape velocity, avoid terms such as 'gravity' or the 'gravitational pull / attraction' of the planet. It is best to refer to its gravitational field. This equation is given on the data sheet, but make sure you know how it is derived.

You've read 0 of your 5 free revision notes this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

    Test yourselfFlashcards

    Did this page help you?

    Previous:Kepler's Laws of Planetary MotionNext:Orbital Motion, Speed & Energy (HL)
    Katie M

    Author: Katie M

    Expertise: Physics

    Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.