Gravitational Potential Energy and Energy Changes in Space

Expert reviewed 22 November 2024 8 minute read


What is Gravitational Potential Energy?

Gravitational potential energy is a fundamental concept in physics that describes the energy an object possesses due to its position within a gravitational field. This concept is crucial for understanding the behaviour of objects in space, planetary motion, and satellite orbits.

Gravitational Potential Energy in Radial Fields

In a radial gravitational field, such as those surrounding planets and stars, the gravitational potential energy (U)(U) of an object is given by:

U=GMmrU = -\frac{GMm}{r}

Where:

  • GG is the universal gravitational constant (6.67×1011Nm2/kg2)(6.67 × 10^{−11}\:Nm^2/kg^2)
  • MM is the mass of the body producing the gravitational field (in kg)
  • mm is the mass of the object in the gravitational field (in kg)
  • rr is the distance between the centres of the two masses (in meters)

Key Points:

  • The negative sign indicates that gravitational potential energy decreases as objects move closer together.
  • Gravitational potential energy is zero at infinite distance and becomes more negative as r decreases.
  • This formula is more accurate than the familiar U=mghU = mgh for large-scale astronomical calculations.

Comparison: Near-Earth vs. Space Applications

AspectNear Earth SurfaceIn Space
FormulaU=mgh*U = mgh*U=GMmrU = -\frac{GMm}{r}
AssumptionConstant gravitational fieldVariable gravitational field
ApplicationSmall displacements near Earth’s surfaceLarge distances in space

Total Energy of Orbiting Bodies

For objects in orbit, the total energy (E)(E) is the sum of kinetic energy (K)(K) and potential energy (U)(U):

E=K+U=GMm2rE = K + U \\= -\frac{GMm}{2r}

This equation shows that the total energy of an orbiting body is negative and equal to half its potential energy.

Practice Question 1

Calculate the total energy of a 1000 kg satellite orbiting Earth at an altitude of 200 km. (Earth’s mass =5.97×1024  kg= 5.97 × 10^{24}\; kg, Earth’s radius = 6371 km)

Calculate orbital radius:

r=6371+200=6571=6.571×106mr = 6371  + 200  \\= 6571  \\= 6.571 × 10^6 m

Apply the total energy formula:

E=GMm2r=(6.67×1011)(5.97×1024)(1000)2(6.571×106)3.03×1010JE = -\frac{GMm}{2r} \\= -\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(1000)}{2(6.571 \times 10^6)} \\ ≈ − 3.03 × 10^{10} J

The negative value indicates that the satellite is bound to Earth’s orbit.

Energy Changes Between Orbits

When a satellite moves between orbits, its total energy changes. The work done to change orbits is equal to the change in total energy:

W=ΔE=EfEi=GMm2rf+GMm2riW = \Delta E \\= E_f - E_i \\= -\frac{GMm}{2r_f} + \frac{GMm}{2r_i}

Where,

  • rfrf and riri are the final and initial orbital radii, respectively.

Practice Question 2

A 500 kg satellite is moved from a circular orbit at an altitude of 300 km to a new circular orbit at 800 km. Calculate the energy change and work done.

Initial radius:

ri=6371+300=6671kmri = 6371 + 300  \\= 6671 \:\text{km}

Final radius:

rf=6371+800=7171  kmr_f = 6371 + 800 \\= 7171 \;\text{km}

Calculate energy change:

ΔE=GMm2rf+GMm2ri=(6.67×1011)(5.97×1024)(500)2(7.171×106)+(6.67×1011)(5.97×1024)(500)2(6.671×106)10.4×108J\Delta E = -\frac{GMm}{2r_f} + \frac{GMm}{2r_i}\\= -\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(500)}{2(7.171 \times 10^6)} + \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(500)}{2(6.671 \times 10^6)}\\ ≈ 10.4 × 10^8 J

The positive value indicates an increase in energy, which makes sense as the satellite is moved to a higher orbit.

Work Done in Gravitational Fields

Work done in a gravitational field is directly related to changes in gravitational potential energy:

W=ΔU=UfUi=GMmrf+GMmriW = \Delta U \\= U_f - U_i \\= -\frac{GMm}{r_f} + \frac{GMm}{r_i}

Work Done by Gravity vs. Work Done Against Gravity

  • Work done by gravity: As an object moves closer to the source of the gravitational field, work is done by gravity, and potential energy decreases (becomes more negative).
  • Work done against gravity: To move an object away from the source of the gravitational field, work must be done against gravity, increasing the object’s potential energy.

Return to Module 5: Advanced Mechanics