Gravitation · Lesson 5 of 7
Escape Speed and Motion Between Gravitating Bodies
“Use energy conservation to find escape thresholds, turning heights and motion across a gravitational barrier.”
• Derive escape speed from conservation of energy. • State the dependence on launch position and the model assumptions. • Calculate height limits and excess speed at infinity. • Apply both force balance and energy to the PDF’s two-source examples.
What does it mean to escape?
A thrown object initially loses speed as gravity raises its potential energy. With insufficient total energy it reaches a maximum distance and returns. The limiting escape case just reaches arbitrarily large distance with speed tending to zero. This is an ideal coasting projectile after launch: neglect air resistance, later engine thrust, Earth’s rotation and the gravity of other bodies unless a problem includes them. The source calls this escape speed; velocity would also require a direction.
Derive the threshold from the energy ledger
Since the square of final speed cannot be negative, the initial total energy must be at least zero for escape. At the threshold, v∞=0. Equating the initial kinetic energy to the magnitude of its negative potential energy gives the result. The projectile mass cancels because both energy terms scale with it. This does not mean the energy required is independent of mass: the required speed is independent, while the corresponding energy is proportional to mass.
For Earth the surface result is about 11.2 km/s. For the Moon the smaller mass-to-radius ratio gives about 2.3 km/s, as rounded in the source. The lower escape speed helps explain why the Moon does not retain a substantial atmosphere: sufficiently fast gas particles can escape. This is a qualitative thermal-escape connection, not a claim that every molecule has the same speed.
Which launch details matter?
In the spherical nonrotating source model, escape speed depends on radial launch distance, not projectile mass or launch direction for a collision-free outward trajectory. Two surface sites at the same radius have the same threshold. A direction aimed through the solid Earth is not a physically available escape route. In a more realistic Earth model, rotation and nonspherical gravity can introduce location and direction considerations; they are outside the calculation developed here. This resolves the source exercise’s wording about launch location: specify whether its radius or the idealised assumptions have changed.
Problem
Source Exercise 7.18 gives ve=11.2 km/s and launch speed vi=3ve. Find speed far from Earth.
- 1.From energy conservation, v∞² = vi²−ve².
- 2.With vi=3ve, v∞=√8 ve = 2√2×11.2 km/s.
- 3.The result is approximately 31.7 km/s. The body has positive total energy rather than merely the zero-energy escape threshold.
A sub-escape launch and its turning height
Problem
Use source Exercise 7.17 values v0=5.0×10³ m/s, M=6.0×10²⁴ kg, R=6.4×10⁶ m and G=6.67×10⁻¹¹ SI. Find maximum distance and altitude.
- 1.GM=4.002×10¹⁴ m³/s².
- 2.1/rmax = 1/(6.4×10⁶) − (5.0×10³)²/(2×4.002×10¹⁴).
- 3.rmax ≈ 7.999×10⁶ m, about 8.0×10⁶ m from Earth’s centre.
- 4.The maximum height above the surface is about 1.60×10⁶ m. The appendix’s 8.0×10⁶ m answer is a centre distance, not an altitude.
A neutral point between two sources
Problem
Source Exercise 7.12 gives Earth–Sun separation D=1.5×10¹¹ m, ME=6×10²⁴ kg and MS=2×10³⁰ kg. Find where their gravitational pulls on a rocket cancel.
- 1.For a point between them at distance x from Earth, equate GME m/x² to GMS m/(D−x)². The projectile mass and G cancel.
- 2.Taking positive square roots gives (D−x)/x = √(MS/ME), so x = D/[1+√(MS/ME)].
- 3.Substitution gives x≈2.59×10⁸ m from Earth, rounded to 2.6×10⁸ m.
- 4.This is an instantaneous zero-gravitational-force point in the fixed-source model. It is not a stable parking orbit or a statement about all forces in a rotating frame.
Problem
Source Example 7.4 fixes spheres of mass M and 4M, each of radius R, with centres 6R apart. A projectile starts at the facing surface of M. Find the critical launch speed toward the other sphere.
- 1.Let r be measured from the centre of M. Between the spheres, force balance gives GM/r² = 4GM/(6R−r)², hence the relevant neutral point is r=2R.
- 2.At launch, the distances are R and 5R, so Ui = −GMm/R−4GMm/(5R) = −9GMm/(5R).
- 3.At the neutral point, the distances are 2R and 4R, giving UN = −3GMm/(2R). The potential increase is 3GMm/(10R).
- 4.Equating ½mvc² to that increase gives vc=√[3GM/(5R)], the source’s limiting threshold. Beyond the neutral point the larger sphere pulls the projectile toward itself.
- 5.Strictly, a projectile with exactly the threshold energy approaches the smooth potential maximum with speed tending to zero; to cross it in finite time requires a speed infinitesimally greater. The textbook threshold is the limiting value.
- 6.At the facing surface of 4M the distances are 5R and R. In the limiting energy calculation, vimpact²=27GM/(5R), so the eventual impact speed is nonzero.
Check your understanding
Quiz
At the limiting escape threshold, speed at infinity tends to
Total energy is zero with U(∞)=0.
Escape speed depends on the projectile’s
The mass cancels from the energy equation.
Launching at 3ve gives v∞ equal to
Subtract ve² from vi².
In the 5 km/s source problem, 8.0×10⁶ m is the maximum
Subtract Earth’s radius to find altitude.
Between fixed M and 4M sources 6R apart, the neutral point lies
Force equality gives 6R−r=2r.
Key Takeaways
• Escape requires nonnegative total energy relative to infinity in the ideal model. • Escape speed is independent of projectile mass, but escape energy is not. • Distinguish radius from altitude and a critical energy from finite-time barrier crossing.