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Gravitational Fields overview

Topic 4 of 5

Escape using an energy account

Minimum escape means reaching indefinitely large separation with speed tending to zero. Set that final condition in an energy account rather than extending constant-g mgh to an infinite height.

Use a spherical source of mass M, treated as fixed, and a much smaller body of mass m starting at centre distance r. Neglect atmosphere, other massive bodies, dissipative effects and propulsion after release. Choose a path that avoids collision with the source; begin with an outward launch.

With zero gravitational potential energy at infinity, UG = -GMm/r. Mechanical energy Ek + UG is conserved during this unpowered motion under the stated assumptions.

Derive the minimum speed from the final condition

For the minimum escape case, both UG and Ek tend to zero at indefinitely large separation. The initial mechanical energy must therefore be zero:

½mvescape2 - GMm/r = 0
½vescape2 = GM/r
vescape = √(2GM/r)

The test mass cancels from the speed threshold. Its required kinetic energy still depends on mass. This equation gives a speed magnitude, not a unique velocity direction; the proposed direction must also avoid the source and satisfy the model assumptions.

Estimate the scale

With rough GM about 4 × 1014 m3/s2 and R about 6 × 106 m, the square root of 2GM/R is of order 104 m/s. For a roughly 400 kg body, the required kinetic energy is of order 3 × 1010 J, matching the magnitude of its initial negative potential energy.

Worked surface escape

Balance positive kinetic and negative potential energy

Use the supplied model GM = 4.0 × 1014 m3/s2 and surface radius R = 6.4 × 106 m:

vescape = √[(2 × 4.0 × 1014)/(6.4 × 106)]
= 1.118 × 104 m/s
≈ 11.2 km/s

For the 400 kg body, the initial potential energy is -2.50 × 1010 J, so the minimum initial kinetic energy is +2.50 × 1010 J.

At the minimum escape threshold, total mechanical energy is zero

Release the 400 kg body outward from the model surface. Assume no atmosphere, no further propulsion, no other significant source and negligible central-body motion.

Positive surface kinetic energy exactly balances negative gravitational potential energyThe vertical energy scale is in ten to the ten joules. At the surface, a positive kinetic-energy bar of 2.50 and a negative gravitational potential-energy bar of minus 2.50 have equal heights, 120 drawing units, on opposite sides of zero. Their total is zero, marked by a point rather than a bar. For minimum escape the body moves outward while kinetic energy decreases and gravitational potential energy increases. The separately written far-separation limit has K tending to zero, U subscript G tending to zero from below and total energy zero. This is a limiting energy account, not an infinity position at a finite point on a trajectory.Energy / 1010 JAt surface release0K+2.50-2.50UGTotal0As r grows without bound:K → 0; UG → 0 from belowTotal energy remains zero.

The surface threshold is K = 2.50 × 1010 J, giving speed about 11.2 km/s. At minimum escape the speed tends to zero only at indefinitely large separation; it does not become zero at a finite height in this model.

At the minimum threshold, positive kinetic energy and negative gravitational potential energy sum to zero. As separation increases without bound, both approach zero. Infinity is a limiting condition, not a finite zero-field location.

From the larger starting radius r = 8.0 × 106 m, the same model gives vescape = 1.00 × 104 m/s = 10.0 km/s. The source is less strongly binding there, so less initial speed is required for escape.

Compare launch speeds using total energy

  • At the minimum speed: total mechanical energy is zero, and speed tends to zero at infinity.
  • Above the minimum: total mechanical energy is positive, leaving nonzero kinetic energy at large separation in this model.
  • Below the minimum: total mechanical energy is negative, so the body cannot reach infinity without further energy input under the stated assumptions.

For a greater launch speed u and residual far-away speed v∞, the account becomes ½mu2 - GMm/r = ½mv∞2. Minimum escape is the special case v∞ = 0.

Real launch energy can involve source rotation, atmospheric losses and continuing propulsion. The ideal threshold does not include those effects. Surface g also decreases with altitude; a constant-g mgh expression cannot describe lifting to infinity.

Optional check For an unpowered outward launch in an isolated, fixed-source gravity model, which final condition defines the minimum escape speed from a given radius?
For an unpowered outward launch in an isolated, fixed-source gravity model, which final condition defines the minimum escape speed from a given radius?