Unit 7 · Topic 04

Orbits and satellites

A circular orbit is curved free fall: gravity supplies the centripetal acceleration.

Unit 7Orbits and satellitesOpen navigation

Concept 01

Centripetal gravity

Essential The minimum you should retain

In a circular orbit GMm/r²=mv²/r and v_orb=sqrt(GM/r).

UnderstandInterpret and connect

Net force is radial and nonzero; velocity is tangent.

DeepenFormulation and conditions

The period is T=2πsqrt(r³/(GM)).

ExploreConnections for further study

A higher circular orbit has lower v and a longer T around the same source.

Worked example

Circular-orbit speed and period

An Earth satellite orbits at r=7.00×10^6 m with μ_E=3.986×10^14 m³/s².

Given
  • r=7.00×10^6 m
  • μ_E=3.986×10^14 m³/s²
Target

Find v_orb and T.

  1. Speed

    v=sqrt(μ/r)≈7546 m/s.

  2. Period

    T=2πsqrt(r³/μ).

  3. Calculation

    T≈5829 s.

  4. Conversion

    T≈97.1 min.

Conclusion

v≈7.55 km/s and T≈97.1 min.

Mathematical relation

Circular orbit

vorb=GMr,T=2πr3GM v_{orb}=\sqrt{GM/r},\quad T=2\pi\sqrt{r^3/(GM)}
Represents

Speed and period of a circular orbit.

Physical interpretation

Gravity is the centripetal force, not zero force.

DeepenVariables, conditions, and checks

Variables

r
orbital radius; usual unit: m
T
orbital period; usual unit: s

Conditions of application

  • Ideal circular orbit around M.

Dimensional check

m/s and s.

A satellite in a circular orbit shows tangential velocity and radial force.vF_gMsatellite

Velocity is tangent and F_g points toward the center: gravity supplies the centripetal acceleration.

Concept 02

A bound state

Essential The minimum you should retain

In a circular orbit K=GMm/(2r), U=−GMm/r, and E=−GMm/(2r).

UnderstandInterpret and connect

The relations are K=−E and U=2E.

DeepenFormulation and conditions

E<0 confirms that the ideal circular orbit is bound.

ExploreConnections for further study

Total energy becomes less negative as radius increases.

Mathematical relation

Circular-orbit energy and escape

K=GMm2r,U=−GMmr,E=−GMm2r,vesc=2GMr K=GMm/(2r),\ U=-GMm/r,\ E=-GMm/(2r),\ v_{esc}=\sqrt{2GM/r}
Represents

Circular-orbit energies and ideal minimum escape speed.

Physical interpretation

At the same r, v_esc=sqrt(2)v_orb and E_circ<0.

DeepenVariables, conditions, and checks

Variables

E
orbital mechanical energy; usual unit: J
v_esc
escape speed; usual unit: m/s

Conditions of application

  • U(∞)=0; ideal model without losses or later propulsion.

Dimensional check

J and m/s.

Signed bars show positive K, negative U, and negative E.K=+1U=−2E=−1zero

For a circular orbit, K:U:E keeps the ratio +1:−2:−1.

Concept 03

Ideally reaching infinity

Essential The minimum you should retain

v_esc=sqrt(2GM/r) is the minimum ideal speed to reach infinity with zero final speed.

UnderstandInterpret and connect

At the same radius v_esc=sqrt(2)v_orb.

DeepenFormulation and conditions

The calculation assumes no later propulsion and no losses.

ExploreConnections for further study

Gravity keeps acting throughout the escape.

Concept 04

Not floating without gravity

Essential The minimum you should retain

Tangential velocity makes the satellite fall continuously without reaching the ideal surface.

UnderstandInterpret and connect

Satellite and occupants share free fall.

DeepenFormulation and conditions

A speed different from circular changes the path and may produce an ellipse or another conic.

ExploreConnections for further study

Orbit does not mean force balance or uniform straight-line motion.

Concept review

Common errors

Each warning includes a concrete way to review the reasoning, not only an incorrect-answer marker.

Claiming net force is zero in a circular orbit.

Nonzero radial force supplies centripetal acceleration.

Treating escape as leaving a gravity-free region.

Gravity acts all the way to infinity; escape speed comes from energy.