Universal Gravitation and Orbits

Newton's law of gravitation, orbital mechanics, and Kepler's laws

Universal Gravitation and Orbits

Newton's Law of Universal Gravitation

F=Gm1m2r2F = G\frac{m_1m_2}{r^2}

where G=6.67×1011G = 6.67 \times 10^{-11} N·m²/kg² is gravitational constant.

Vector form: F12=Gm1m2r2r^12\vec{F}_{12} = -G\frac{m_1m_2}{r^2}\hat{r}_{12}

Gravitational Potential Energy

Choosing U=0U = 0 at r=r = \infty:

U(r)=Gm1m2rU(r) = -G\frac{m_1m_2}{r}

Work to bring masses from infinity to separation rr:

W=rFdr=Gm1m2rW = \int_\infty^r F \, dr' = -G\frac{m_1m_2}{r}

Gravitational Field

g=GMr2r^\vec{g} = -\frac{GM}{r^2}\hat{r}

Gravitational potential: Φ=GMr\Phi = -\frac{GM}{r}

g=Φ\vec{g} = -\nabla\Phi

Escape Velocity

Minimum velocity to escape gravitational field:

Set total energy = 0: 12mvesc2GMmR=0\frac{1}{2}mv_{esc}^2 - G\frac{Mm}{R} = 0

vesc=2GMRv_{esc} = \sqrt{\frac{2GM}{R}}

For Earth: vesc11.2v_{esc} \approx 11.2 km/s

Circular Orbits

For circular orbit of radius rr:

Centripetal force = Gravitational force: mv2r=GMmr2\frac{mv^2}{r} = G\frac{Mm}{r^2}

Orbital velocity: v=GMrv = \sqrt{\frac{GM}{r}}

Orbital period: T=2πrv=2πr3GMT = \frac{2\pi r}{v} = 2\pi\sqrt{\frac{r^3}{GM}}

Orbital energy: E=KE+PE=12mv2GMmrE = KE + PE = \frac{1}{2}mv^2 - G\frac{Mm}{r}

E=GMm2rE = -\frac{GMm}{2r}

(Negative: bound orbit)

Note: E=KE|E| = KE (virial theorem)

Kepler's Laws

First Law (Law of Ellipses): Planets move in elliptical orbits with the Sun at one focus.

Second Law (Law of Equal Areas): Line from Sun to planet sweeps equal areas in equal times.

This follows from angular momentum conservation: dAdt=L2m=constant\frac{dA}{dt} = \frac{L}{2m} = \text{constant}

Third Law (Harmonic Law): T2a3T^2 \propto a^3

where aa is semi-major axis.

For circular orbit (a=ra = r): T2=4π2GMr3T^2 = \frac{4\pi^2}{GM}r^3

Elliptical Orbits

Energy: E=GMm2aE = -\frac{GMm}{2a}

where aa is semi-major axis.

Semi-major axis from energy: a=GMm2Ea = -\frac{GMm}{2E}

Angular momentum: L=mGMa(1e2)L = m\sqrt{GMa(1-e^2)}

where ee is eccentricity.

Eccentricity: e=1+2EL2m(GM)2e = \sqrt{1 + \frac{2EL^2}{m(GM)^2}}

Perihelion and Aphelion

Perihelion (closest): rp=a(1e)r_p = a(1-e)

Aphelion (farthest): ra=a(1+e)r_a = a(1+e)

Velocities: vp=GMa1+e1e,va=GMa1e1+ev_p = \sqrt{\frac{GM}{a}\frac{1+e}{1-e}}, \quad v_a = \sqrt{\frac{GM}{a}\frac{1-e}{1+e}}

(From energy and angular momentum conservation)

Hohmann Transfer Orbit

Efficient orbit change between circular orbits:

Transfer orbit energy: Et=GMmr1+r2E_t = -\frac{GM m}{r_1 + r_2}

Velocity changes: Δv1=GMr1(2r2r1+r21)\Delta v_1 = \sqrt{\frac{GM}{r_1}}\left(\sqrt{\frac{2r_2}{r_1+r_2}} - 1\right)

Δv2=GMr2(12r1r1+r2)\Delta v_2 = \sqrt{\frac{GM}{r_2}}\left(1 - \sqrt{\frac{2r_1}{r_1+r_2}}\right)

Reduced Mass Problem

Two-body problem reduces to one-body with reduced mass:

μ=m1m2m1+m2\mu = \frac{m_1m_2}{m_1+m_2}

Both orbit common center of mass.

Gravitational Force Inside Sphere

For uniform sphere, only mass at r<rr' < r contributes:

Menclosed=Mr3R3M_{enclosed} = M\frac{r^3}{R^3}

F(r)=GMmR3rF(r) = G\frac{Mm}{R^3}r

(Linear with rr, like spring force!)

At center: F=0F = 0

Shell Theorem

  1. Uniform spherical shell exerts no force on particle inside
  2. Shell acts as point mass for particle outside

These allow us to treat planets as point masses for external objects.

📚 Practice Problems

1Problem 1easy

Question:

Calculate the gravitational force between Earth (M_E = 5.97 × 10²⁴ kg) and the Moon (M_M = 7.35 × 10²² kg), separated by r = 3.84 × 10⁸ m. Use G = 6.67 × 10⁻¹¹ N·m²/kg².

💡 Show Solution

Given:

  • M_E = 5.97 × 10²⁴ kg
  • M_M = 7.35 × 10²² kg
  • r = 3.84 × 10⁸ m
  • G = 6.67 × 10⁻¹¹ N·m²/kg²

Gravitational force:

F=GMEMMr2F = \frac{GM_E M_M}{r^2}

F=(6.67×1011)(5.97×1024)(7.35×1022)(3.84×108)2F = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(7.35 \times 10^{22})}{(3.84 \times 10^8)^2}

Numerator: (6.67)(5.97)(7.35)×1011+24+22=292.5×1035(6.67)(5.97)(7.35) \times 10^{-11+24+22} = 292.5 \times 10^{35}

Denominator: (3.84)2×1016=14.75×1016(3.84)^2 \times 10^{16} = 14.75 \times 10^{16}

F=292.5×103514.75×1016=19.8×1019F = \frac{292.5 \times 10^{35}}{14.75 \times 10^{16}} = 19.8 \times 10^{19}

F=1.98×1020 N\boxed{F = 1.98 \times 10^{20} \text{ N}}

This is the force keeping the Moon in orbit!

2Problem 2medium

Question:

A satellite orbits Earth at altitude h = 400 km above the surface. Given R_E = 6.37 × 10⁶ m and g = 9.8 m/s² at surface, find: (a) the orbital speed, (b) the orbital period, and (c) the satellite's acceleration.

💡 Show Solution

Given:

  • h = 400 km = 4.0 × 10⁵ m
  • R_E = 6.37 × 10⁶ m
  • g = 9.8 m/s²

Orbital radius: r=RE+h=6.77×106r = R_E + h = 6.77 \times 10^6 m

(a) Orbital speed:

For circular orbit: mv2r=GMEmr2\frac{mv^2}{r} = \frac{GM_E m}{r^2}

v=GMErv = \sqrt{\frac{GM_E}{r}}

Using g=GMERE2g = \frac{GM_E}{R_E^2}GME=gRE2GM_E = gR_E^2:

v=gRE2r=gRErv = \sqrt{\frac{gR_E^2}{r}} = \sqrt{g}\frac{R_E}{\sqrt{r}}

v=9.86.37×1066.77×106v = \sqrt{9.8} \cdot \frac{6.37 \times 10^6}{\sqrt{6.77 \times 10^6}}

v=3.13×6.37×1062.60×103v = 3.13 \times \frac{6.37 \times 10^6}{2.60 \times 10^3}

v=7670 m/s=7.67 km/s\boxed{v = 7670 \text{ m/s} = 7.67 \text{ km/s}}

(b) Orbital period:

T=2πrv=2π(6.77×106)7670T = \frac{2\pi r}{v} = \frac{2\pi(6.77 \times 10^6)}{7670}

T=4.25×1077670=5546 sT = \frac{4.25 \times 10^7}{7670} = 5546 \text{ s}

T=92.4 min=1.54 hours\boxed{T = 92.4 \text{ min} = 1.54 \text{ hours}}

(c) Acceleration:

Centripetal acceleration: a=v2r=(7670)26.77×106a = \frac{v^2}{r} = \frac{(7670)^2}{6.77 \times 10^6}

a=5.88×1076.77×106a = \frac{5.88 \times 10^7}{6.77 \times 10^6}

a=8.69 m/s2\boxed{a = 8.69 \text{ m/s}^2}

This is 89% of g at surface (slightly weaker due to altitude).

3Problem 3hard

Question:

Derive the escape velocity from Earth's surface using energy conservation. Then calculate the numerical value for Earth (M_E = 5.97 × 10²⁴ kg, R_E = 6.37 × 10⁶ m). Also find the escape velocity from the Moon (M_M = 7.35 × 10²² kg, R_M = 1.74 × 10⁶ m).

💡 Show Solution

Derivation:

At surface: Ei=KE+PE=12mv2GMEmREE_i = KE + PE = \frac{1}{2}mv^2 - \frac{GM_E m}{R_E}

At infinity (just escaping): Ef=0+0=0E_f = 0 + 0 = 0

Energy conservation: Ei=EfE_i = E_f

12mvesc2GMEmRE=0\frac{1}{2}mv_{esc}^2 - \frac{GM_E m}{R_E} = 0

vesc=2GMEREv_{esc} = \sqrt{\frac{2GM_E}{R_E}}

Using g=GME/RE2g = GM_E/R_E^2:

vesc=2gRE\boxed{v_{esc} = \sqrt{2gR_E}}

From Earth:

vesc=2(9.8)(6.37×106)v_{esc} = \sqrt{2(9.8)(6.37 \times 10^6)}

vesc=1.25×108=1.12×104v_{esc} = \sqrt{1.25 \times 10^8} = 1.12 \times 10^4

vesc,Earth=11,200 m/s=11.2 km/s\boxed{v_{esc,Earth} = 11,200 \text{ m/s} = 11.2 \text{ km/s}}

From Moon:

Surface gravity on Moon: gM=GMMRM2=(6.67×1011)(7.35×1022)(1.74×106)2g_M = \frac{GM_M}{R_M^2} = \frac{(6.67 \times 10^{-11})(7.35 \times 10^{22})}{(1.74 \times 10^6)^2}

gM=1.62 m/s2g_M = 1.62 \text{ m/s}^2

vesc,Moon=2(1.62)(1.74×106)v_{esc,Moon} = \sqrt{2(1.62)(1.74 \times 10^6)}

vesc,Moon=5.64×106v_{esc,Moon} = \sqrt{5.64 \times 10^6}

vesc,Moon=2370 m/s=2.37 km/s\boxed{v_{esc,Moon} = 2370 \text{ m/s} = 2.37 \text{ km/s}}

Moon's lower mass and smaller radius make escape much easier!

Fun fact: This is why Moon has no atmosphere - gas molecules can reach v_esc and escape.