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Topic 4 of 11

Central Force Motion

When the force between two bodies depends only on their separation — as with gravity and electrostatics — the problem reduces to one-dimensional motion in an effective potential. This yields Kepler's laws as exact consequences of Newton's inverse-square force.

Key Concepts

Reduced Mass

For two bodies

m_1,m_2

, the two-body problem is equivalent to a one-body problem with reduced mass

\mu=m_1m_2/(m_1+m_2)

moving in the central potential. Angular momentum

l=\mu r^2\dot\phi

is conserved.

Effective Potential

V_{\rm eff}(r)=V(r)+l^2/(2\mu r^2)

. The centrifugal term

l^2/(2\mu r^2)

creates a potential barrier. Bound orbits oscillate between turning points where

E=V_{\rm eff}

Kepler's Laws

(1) Orbits are conic sections (ellipses for bound orbits). (2) Equal areas swept in equal times (conservation of

l

). (3)

T^2\propto a^3

— the square of the period is proportional to the cube of the semi-major axis.

Vis-Viva Equation

For any point on a Keplerian orbit:

v^2=GM(2/r-1/a)

, where

a

is the semi-major axis. Gives speed at any orbital radius directly.

Key Equations

Kepler's Third Law

T^2 = \frac{4\pi^2}{GM}a^3

Period squared proportional to semi-major axis cubed. For solar orbits, T²(yr) = a³(AU).

Vis-viva

v^2 = GM\left(\frac{2}{r} - \frac{1}{a}\right)

Speed at any point in a Keplerian orbit; a = semi-major axis.

Escape velocity

v_{\rm esc} = \sqrt{\frac{2GM}{R}}

Minimum speed to escape from the surface (radius R) of a body of mass M.

Worked Example

Orbital Speed of a Low Earth Orbit Satellite

Problem

A satellite orbits Earth in a circular orbit at altitude $h=400$ km. Find its orbital speed. ( $GM_E=3.986\times10^{14}$ m³/s², $R_E=6.371\times10^6$ m.)

Solution

Orbital radius: $r=R_E+h=6.371\times10^6+4.0\times10^5=6.771\times10^6$ m.

For circular orbit, gravity provides centripetal force: $GM/r^2=v^2/r$ .

v = \sqrt{\frac{GM_E}{r}} = \sqrt{\frac{3.986\times10^{14}}{6.771\times10^6}}

v = \sqrt{5.887\times10^7} = 7673\text{ m/s} \approx 7.67\text{ km/s}

Answer v ≈ 7670 m/s (7.67 km/s). The ISS orbits at about this speed.

Practice

Exercises

7 problems

1 of 7

Using Kepler's third law with Earth's orbit as reference (T=1 yr, a=1 AU), find Mars's orbital period (in yr) given $a_{\rm Mars}=1.524$ AU.

Your answer

2 of 7

Find the circular orbital speed (in m/s) of a satellite at $r=6.8\times10^6$ m from Earth's center. ( $GM_E=3.986\times10^{14}$ m³/s²)

Your answer

m/s

3 of 7

Find Earth's escape velocity (in m/s) at the surface. ( $GM_E=3.986\times10^{14}$ m³/s², $R_E=6.371\times10^6$ m.)

Unlock Exercise 3