Gravitational potential energy $E_{p} = mV = -GMm/r$ for a mass m at distance r from mass M
Gravitational Fields - OCR A-Level Physics
- Gravitational potential energyThe capacity to do work. Measured in joules (J).The energyThe capacity to do work. Measured in joules (J). an object possesses due to its position in a gravitational fieldA region of space in which a mass experiences a gravitational force.. $E_{p} = mV = -GMm/r$ for a mass m at distance r from mass M.
- The change in GPE when moving from r_1 to r_2: $\Delta$E_p = -GMm(1/r_2 - 1/r_1).
- Escape velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.The minimum speed at which an object must be launched from the surface of a body to escape its gravitational fieldA region of space in which a mass experiences a gravitational force. entirely (reach infinity with zero kinetic energyThe capacity to do work. Measured in joules (J).The energy an object possesses due to its motion.). is the minimum speed needed for an object to escape the gravitational fieldA region of space in which a mass experiences a gravitational force. of a planet (reach infinity with zero KE).
- Derived from energy conservation: $\frac{1}{2}$mv^2 + (-GMm/r) = 0 (KE at surface + PE at surface = total energy at infinity = 0).
- This gives $v_{escape} =$ $\sqrt{2GM/r} =$ $\sqrt{2gr}$where r is the radius of the planet.
- Escape velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.The minimum speed at which an object must be launched from the surface of a body to escape its gravitational field entirely (reach infinity with zero kinetic energyThe energy an object possesses due to its motion.). does not depend on the mass of the escaping object.
- For Earth: $v_{escape} \approx$ 11.2 km $s^{-1}$.
Worked Example
Calculate the escape velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.The minimum speed at which an object must be launched from the surface of a body to escape its gravitational field entirely (reach infinity with zero kinetic energyThe energy an object possesses due to its motion.). from Earth. M_E = $5.97 \times 10^{24}$ kg, R_E = $6.37 \times $\(10^{6}\)$ m.
Show Solution
1
$v = \sqrt{2GM/R}.$
2
$v = \sqrt{2 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24} / (6.37 \times 10^6)}.$
3
$v = \sqrt{(7.96 \times 10^{14}) / (6.37 \times 10^6)}.$
4
$v = \sqrt{1.25 \times 10^{8}}.$
5
$v = 1.12 \times 10^4 m s^{-1} = 11.2 km s^{-1}.$
Answer
v_{escape} = 11.2 km $s^{-1}$
Common Mistake
MEDIUM
Wrong: Using mgh for gravitational PE at large distances from a planet.
Right: mgh only works near the surface where g is approximately constant. For large distances, use $E_{p} = -GMm/r. The mgh$ formula is a linear approximation of the true 1/r relationship.
Right: mgh only works near the surface where g is approximately constant. For large distances, use $E_{p} = -GMm/r. The mgh$ formula is a linear approximation of the true 1/r relationship.