Gravitation · Lesson 6 of 7
Satellites: Orbits, Energy and Weightlessness
“Keep circular-orbit dynamics, orbital energy, mass estimates and free-fall effects in one connected lesson.”
• Derive circular orbital speed and period from gravitational force. • Use orbital observations to estimate masses. • Calculate satellite kinetic, potential and total energy. • Explain orbital transfers, escape energy and weightlessness.
An orbit is continuous gravitational free fall
A satellite has sideways motion while gravity continually bends its trajectory. In a circular orbit, its speed is constant but velocity direction changes, so there is inward acceleration. Gravity supplies that acceleration; “centripetal force” is the role of the net inward force, not an extra force added beside gravity. The source treats the satellite mass as small compared with Earth and ignores drag and other bodies for the circular-orbit derivation.
At a common radius, the escape threshold is √2 times circular speed. A satellite already has kinetic energy, so it needs less additional energy to escape than a body held stationary at the same radius. This comparison concerns energy addition; the exact trajectory produced also depends on how a propulsion system changes the velocity.
Orbital period and the period law
A larger circular orbit is longer and is traversed at a lower speed, so the period increases. For an orbit idealised as just above Earth’s surface, use r≈RE and GM=g0RE². Then T0=2π√(RE/g0), approximately 85 minutes with RE=6400 km and g0=9.8 m/s². This is a mathematical near-surface estimate, not a practical orbit through air or terrain.
Weighing a central body with its satellite
Problem
Source Example 7.5 gives Phobos period 7 h 39 min and circular orbit radius 9.4×10³ km. It also gives Mars–Sun orbital radius 1.52 times Earth’s. Find Mars mass and the year length.
- 1.Convert Phobos data: T=(7×60+39)×60 = 27540 s and r=9.4×10⁶ m.
- 2.M=4π²r³/(GT²) ≈ 6.48×10²³ kg using the source’s rounded π=3.14 and G=6.67×10⁻¹¹. More digits of π give about 6.48×10²³ kg as well at this precision.
- 3.For planets around the Sun, TM/TE=(aM/aE)^(3/2).
- 4.TM=365×1.52^(3/2) ≈ 684 days. The period constant for Phobos around Mars is not the Sun-orbit constant; only the planetary ratio uses the shared solar constant.
Problem
Complete source Example 7.6 with lunar radius 3.84×10⁸ m and period 27.3 days. Then perform source Example 7.7’s conversion of k=10⁻¹³ s²/m³ into day²/km³.
- 1.Use T=27.3×86400 s and ME=4π²r³/(GT²), obtaining about 6.02×10²⁴ kg with the rounded orbital data. This agrees within about 1% with the surface-gravity estimate 5.97×10²⁴ kg.
- 2.The difference reflects the approximate supplied observations and ideal orbital model; it is not removed by changing the test mass.
- 3.To convert k, replace each second by 1/86400 day and each metre by 1/1000 km. Thus k=10⁻¹³×10⁹/86400² ≈ 1.34×10⁻¹⁴ day²/km³.
- 4.Using the source’s rounded k≈1.33×10⁻¹⁴ with r=3.84×10⁵ km gives T≈27.4 days, about 27.3 days at the textbook’s loose approximation. Using the more exact conversion of the stipulated k=10⁻¹³ gives about 27.5 days.
- 5.The input k itself is rounded; do not claim that substituting it recovers an exact lunar period. Units and the approximate accuracy both matter.
Problem
Use the source Exercises 7.4, 7.13 and 7.5 to estimate astronomical masses and a period.
- 1.For Io, r=4.22×10⁸ m and T=1.769×86400 s. Jupiter’s mass 4π²r³/(GT²) is about 1.90×10²⁷ kg.
- 2.For Earth’s solar orbit, r=1.5×10¹¹ m and T=365×86400 s give MS≈2.01×10³⁰ kg. Thus Jupiter’s mass is about 9.5×10⁻⁴ of the Sun’s, roughly one thousandth.
- 3.For the source’s simplified galaxy, total enclosed mass is taken as 2.5×10¹¹ solar masses and the radius as 50000 light years. Using 1 ly≈9.46×10¹⁵ m gives T=2π√(r³/GM)≈3.55×10⁸ years.
- 4.The galactic result uses the exercise’s idealised spherical/enclosed-mass estimate. A real non-spherical galaxy is not automatically a point mass; the computation is an application of the specified simplified model.
The three orbital energies
Derive K by substituting vc²=GM/r into ½mvc². Add the negative potential energy. The total is negative, meaning that the ideal two-body orbit is bound. For an ellipse, K and U vary with position and the instantaneous relation U=−2K is not generally true. Angular momentum and total mechanical energy remain constant. For a comet in a strongly elliptical orbit, speed, angular speed, kinetic energy and potential energy all vary, while those two conserved quantities do not.
Problem
Source Example 7.8 uses m=400 kg, g0=9.81 m/s² and RE=6.37×10⁶ m. Find the endpoint changes in total, kinetic and potential energy.
- 1.Initial and final total energies are Ei=−GMm/(4RE) and Ef=−GMm/(8RE).
- 2.ΔE=+GMm/(8RE)=+mg0RE/8≈+3.12×10⁹ J, or about +3.13×10⁹ J at the source’s rounding.
- 3.The kinetic energy decreases by the same magnitude: ΔK≈−3.12×10⁹ J.
- 4.Potential energy increases by twice that magnitude: ΔU=+GMm/(4RE)≈+6.25×10⁹ J. Check that ΔK+ΔU=ΔE.
- 5.The source prints a negative sign for ΔU in this example. The energy definitions and direct endpoint subtraction require the positive sign used here.
- 6.These are differences between the final circular states. They do not specify the burns or trajectory used to transfer between them.
Problem
Source Exercise 7.19 has m=200 kg, h=400 km, ME=6.0×10²⁴ kg, RE=6.4×10⁶ m and G=6.67×10⁻¹¹ SI. Find the minimum added energy.
- 1.The centre distance is 6.8×10⁶ m. Initial circular-orbit total energy is −GMm/(2r).
- 2.The escape threshold has E=0, so added energy is GMm/(2r).
- 3.Substitution gives about 5.89×10⁹ J. A body initially stationary at that radius would need GMm/r, twice this energy, in the same ideal model.
Weightlessness without absence of gravity
An astronaut and spacecraft in the same small orbiting region both accelerate in gravitational free fall. The cabin floor does not need to support the astronaut against gravity, so the support-force or scale reading can be zero. Gravitational attraction is still present and supplies their orbital acceleration. Apparent weightlessness is therefore different from zero gravitational field.
A large spacecraft can reveal small differences in gravitational acceleration across its extent: separated objects do not all share exactly the same gravitational acceleration. These differential effects answer the source question about detecting gravity in a large orbiting station. The exercise also associates weightlessness with fluid redistribution toward the head and loss of usual up–down cues, making a swollen face, headache and orientational difficulty its selected possibilities rather than swollen feet. The teaching point is altered support and orientation cues, not disappearance of gravity.
Check your understanding
Quiz
What provides centripetal force for an ideal circular satellite?
Centripetal describes the inward role of gravity.
Increasing circular orbital radius makes the period
T is proportional to r^(3/2).
For a circular gravitational orbit with U(∞)=0, total energy is
U=−2K and E=K+U=−K.
Moving to a higher circular orbit makes U
U=−GMm/r increases toward zero as r grows.
An astronaut is apparently weightless mainly because
Apparent weight measures supporting contact force, not the existence of gravity.
Key Takeaways
• Gravity supplies orbital curvature; it is not accompanied by a separate centripetal interaction. • For circular orbits, vc=√(GM/r), T²∝r³ and E=−GMm/(2r). • A higher circular orbit has lower K but higher U and total E. • Weightlessness is free fall with little or no supporting force.