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Gravitational potential energy and escape velocity explain how objects move near planets, moons, and stars. This cheat sheet helps students connect Newton's law of gravitation to energy conservation. It is useful for solving problems involving satellites, rockets, falling objects, and changes in orbital distance.

Worked-example style formulas make it easier to choose the correct equation for each situation.

The most important idea is that gravitational potential energy depends on distance from the center of a mass, not height above the surface alone. Near Earth, U=mghU = mgh is an approximation, but for large distances the correct formula is U=GMmrU = -\frac{GMm}{r}. Escape velocity comes from setting total mechanical energy equal to zero, giving ve=2GMrv_e = \sqrt{\frac{2GM}{r}}.

Conservation of energy links speed, position, kinetic energy, and gravitational potential energy in one equation.

Key Facts

  • Universal gravitational potential energy is U=GMmrU = -\frac{GMm}{r}, where rr is the center-to-center distance between the two masses.
  • Near a planet's surface, the approximation ΔU=mgΔh\Delta U = mg\Delta h works only when Δh\Delta h is small compared with the planet's radius.
  • The change in gravitational potential energy is ΔU=UfUi=GMmrf+GMmri\Delta U = U_f - U_i = -\frac{GMm}{r_f} + \frac{GMm}{r_i}.
  • Total mechanical energy in a gravitational field is E=K+U=12mv2GMmrE = K + U = \frac{1}{2}mv^2 - \frac{GMm}{r}.
  • Escape speed from distance rr is ve=2GMrv_e = \sqrt{\frac{2GM}{r}}, and it does not depend on the escaping object's mass mm.
  • For a circular orbit, orbital speed is v=GMrv = \sqrt{\frac{GM}{r}}, which is smaller than escape speed by a factor of 2\sqrt{2}.
  • An object escapes if its total mechanical energy is E0E \ge 0, and it remains gravitationally bound if E<0E < 0.
  • The gravitational field strength at distance rr is g=GMr2g = \frac{GM}{r^2}, so gg decreases as distance from the center increases.

Vocabulary

Gravitational Potential Energy
The energy an object has because of its position in a gravitational field, given by U=GMmrU = -\frac{GMm}{r} for two point masses or spherical bodies.
Escape Velocity
The minimum launch speed needed for an object to reach infinitely far away with no additional propulsion, given by ve=2GMrv_e = \sqrt{\frac{2GM}{r}}.
Mechanical Energy
The sum of kinetic energy and potential energy, written as E=K+UE = K + U.
Kinetic Energy
The energy of motion of an object, calculated with K=12mv2K = \frac{1}{2}mv^2.
Orbital Radius
The distance rr from the center of the central body to the orbiting object, not the distance above the surface.
Bound Orbit
A gravitational motion with total mechanical energy E<0E < 0, meaning the object does not escape to infinity.

Common Mistakes to Avoid

  • Using height above the surface for rr is wrong because rr must be measured from the center of the planet or star. Use r=R+hr = R + h when an object is at altitude hh above a body of radius RR.
  • Dropping the negative sign in U=GMmrU = -\frac{GMm}{r} is wrong because gravitational potential energy is defined as zero at infinity and negative at finite distance. The negative sign shows the object is bound to the gravitational source.
  • Using mghmgh for satellite or escape problems is wrong when the distance change is large. Use U=GMmrU = -\frac{GMm}{r} and energy conservation instead.
  • Assuming escape velocity depends on the rocket's mass is wrong because the mass mm cancels when 12mve2=GMmr\frac{1}{2}mv_e^2 = \frac{GMm}{r}. A heavier object needs more energy, but not a higher escape speed.
  • Confusing circular orbital speed with escape speed is wrong because v=GMrv = \sqrt{\frac{GM}{r}} keeps an object in circular orbit, while ve=2GMrv_e = \sqrt{\frac{2GM}{r}} lets it escape.

Practice Questions

  1. 1 A 1200kg1200\,\text{kg} satellite is at a distance r=7.0×106mr = 7.0 \times 10^6\,\text{m} from Earth's center. Using G=6.67×1011Nm2/kg2G = 6.67 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2 and ME=5.97×1024kgM_E = 5.97 \times 10^{24}\,\text{kg}, calculate U=GMEmrU = -\frac{GM_E m}{r}.
  2. 2 Find the escape speed from Earth's surface using ve=2GMEREv_e = \sqrt{\frac{2GM_E}{R_E}}, ME=5.97×1024kgM_E = 5.97 \times 10^{24}\,\text{kg}, and RE=6.37×106mR_E = 6.37 \times 10^6\,\text{m}.
  3. 3 A spacecraft moves from ri=8.0×106mr_i = 8.0 \times 10^6\,\text{m} to rf=1.2×107mr_f = 1.2 \times 10^7\,\text{m} from Earth's center. Write and evaluate ΔU=GMEmrf+GMEmri\Delta U = -\frac{GM_E m}{r_f} + \frac{GM_E m}{r_i} for m=500kgm = 500\,\text{kg}.
  4. 4 Explain why an object can have negative gravitational potential energy but positive kinetic energy, and how the sign of E=K+UE = K + U tells whether it can escape.

Understanding Gravitational Potential Energy and Escape Velocity Worked Examples

The negative sign in gravitational potential energy has a physical meaning. Physicists choose zero potential energy at an infinite distance from the planet or star. An object near Earth is in a bound state because energy must be supplied to separate it completely from Earth.

Moving outward makes the potential energy less negative, so the potential energy increases. Moving inward makes it more negative. This sign is a common source of errors.

A negative potential energy does not mean the object has negative energy in every sense. It means the selected zero level is far away.

In a worked problem, start by marking the initial position and final position. Measure each distance from the planet's center. Then decide whether mechanical energy is conserved.

It is conserved only when gravity is the only force doing significant work. If a rocket engine fires, include the energy added by the engine. If air resistance matters, some mechanical energy becomes thermal energy.

For a freely moving object, compare the kinetic energy plus gravitational potential energy at both locations. This approach can find an unknown speed, maximum height, or minimum launch speed without following every moment of the motion.

Orbit problems require careful thinking because an orbiting object is always falling toward the central body. Its sideways speed keeps it from hitting the surface. A circular orbit has one fixed radius and one fixed speed.

A faster object at that same location does not simply move into a higher circular orbit. It enters a stretched orbit if its speed is below escape speed. To move a satellite into a larger orbit, a spacecraft usually fires its engine forward.

That first burn raises the far side of the orbit. The spacecraft then slows as it climbs, even though the first burn increased its total energy. This result can feel backwards until energy and orbit shape are considered together.

Escape speed describes a special limiting path. An object launched at exactly that speed, with no further pushes or drag, keeps slowing down and approaches zero speed only at an infinite distance. It does not need to keep its initial speed all the way out.

A speed greater than escape speed gives the object positive total mechanical energy and leaves it with some speed far away. In real launches, air drag, fuel use, planetary rotation, and the gravity of other bodies change the details. Space agencies therefore plan a route and a sequence of engine burns rather than treating escape speed as one simple launch target.

When checking answers, look first at the direction of change. Raising an object should require positive work against gravity. A satellite farther from Earth should have a lower circular orbital speed.

Units provide another useful check. Energy is measured in joules, speed in meters per second, distance in meters, and mass in kilograms. Keep the planet mass separate from the object mass.

The object mass cancels from an ideal escape-speed calculation because gravity pulls harder on a heavier object by exactly the same factor that its inertia increases. That cancellation applies only in the ideal two-body model.