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=6.67ร—10โˆ’11G = 6.67 \times 10^{-11} Nยทmยฒ/kgยฒ is gravitational constant.

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

Gravitational Potential Energy

Choosing U=0U = 0 at r=โˆžr = \infty:

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

Work to bring masses from infinity to separation rr:

W=โˆซโˆžrFโ€‰drโ€ฒ=โˆ’Gm1m2rW = \int_\infty^r F \, dr' = -G\frac{m_1m_2}{r}

Gravitational Field

gโƒ—=โˆ’GMr2r^\vec{g} = -\frac{GM}{r^2}\hat{r}

Gravitational potential: ฮฆ=โˆ’GMr\Phi = -\frac{GM}{r}

gโƒ—=โˆ’โˆ‡ฮฆ\vec{g} = -\nabla\Phi

Escape Velocity

Minimum velocity to escape gravitational field:

Set total energy = 0: 12mvesc2โˆ’GMmR=0\frac{1}{2}mv_{esc}^2 - G\frac{Mm}{R} = 0

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

For Earth: vescโ‰ˆ11.2v_{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=2ฯ€rv=2ฯ€r3GMT = \frac{2\pi r}{v} = 2\pi\sqrt{\frac{r^3}{GM}}

Orbital energy: E=KE+PE=12mv2โˆ’GMmrE = KE + PE = \frac{1}{2}mv^2 - G\frac{Mm}{r}

E=โˆ’GMm2rE = -\frac{GMm}{2r}

(Negative: bound orbit)

Note: โˆฃEโˆฃ=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): T2โˆa3T^2 \propto a^3

where aa is semi-major axis.

For circular orbit (a=ra = r): T2=4ฯ€2GMr3T^2 = \frac{4\pi^2}{GM}r^3

Elliptical Orbits

Energy: E=โˆ’GMm2aE = -\frac{GMm}{2a}

where aa is semi-major axis.

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

Angular momentum: L=mGMa(1โˆ’e2)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(1โˆ’e)r_p = a(1-e)

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

Velocities: vp=GMa1+e1โˆ’e,va=GMa1โˆ’e1+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=โˆ’GMmr1+r2E_t = -\frac{GM m}{r_1 + r_2}

Velocity changes: ฮ”v1=GMr1(2r2r1+r2โˆ’1)\Delta v_1 = \sqrt{\frac{GM}{r_1}}\left(\sqrt{\frac{2r_2}{r_1+r_2}} - 1\right)

ฮ”v2=GMr2(1โˆ’2r1r1+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:

ฮผ=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 rโ€ฒ<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.

๐Ÿ“š Practice Problems

1Problem 1easy

โ“ Question:

Calculate the gravitational force between Earth (M_E = 5.97 ร— 10ยฒโด kg) and the Moon (M_M = 7.35 ร— 10ยฒยฒ kg), separated by r = 3.84 ร— 10โธ m. Use G = 6.67 ร— 10โปยนยน Nยทmยฒ/kgยฒ.

๐Ÿ’ก Show Solution

Given:

  • M_E = 5.97 ร— 10ยฒโด kg
  • M_M = 7.35 ร— 10ยฒยฒ kg
  • r = 3.84 ร— 10โธ m
  • G = 6.67 ร— 10โปยนยน Nยทmยฒ/kgยฒ

Gravitational force:

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

F=(6.67ร—10โˆ’11)(5.97ร—1024)(7.35ร—1022)(3.84ร—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)ร—10โˆ’11+24+22=292.5ร—1035(6.67)(5.97)(7.35) \times 10^{-11+24+22} = 292.5 \times 10^{35}

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

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

F=1.98ร—1020ย N\boxed{F = 1.98 \times 10^{20} \text{ N}}

This is the force keeping the Moon in orbit!

2Problem 2medium

โ“ Question:

A satellite orbits Earth at altitude h = 400 km above the surface. Given R_E = 6.37 ร— 10โถ m and g = 9.8 m/sยฒ at surface, find: (a) the orbital speed, (b) the orbital period, and (c) the satellite's acceleration.

๐Ÿ’ก Show Solution

Given:

  • h = 400 km = 4.0 ร— 10โต m
  • R_E = 6.37 ร— 10โถ m
  • g = 9.8 m/sยฒ

Orbital radius: r=RE+h=6.77ร—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} โ†’ GME=gRE2GM_E = gR_E^2:

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

v=9.8โ‹…6.37ร—1066.77ร—106v = \sqrt{9.8} \cdot \frac{6.37 \times 10^6}{\sqrt{6.77 \times 10^6}}

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

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

(b) Orbital period:

T=2ฯ€rv=2ฯ€(6.77ร—106)7670T = \frac{2\pi r}{v} = \frac{2\pi(6.77 \times 10^6)}{7670}

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

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

(c) Acceleration:

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

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

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

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

3Problem 3hard

โ“ Question:

Derive the escape velocity from Earth's surface using energy conservation. Then calculate the numerical value for Earth (M_E = 5.97 ร— 10ยฒโด kg, R_E = 6.37 ร— 10โถ m). Also find the escape velocity from the Moon (M_M = 7.35 ร— 10ยฒยฒ kg, R_M = 1.74 ร— 10โถ m).

๐Ÿ’ก Show Solution

Derivation:

At surface: Ei=KE+PE=12mv2โˆ’GMEmREE_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

12mvesc2โˆ’GMEmRE=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)(6.37ร—106)v_{esc} = \sqrt{2(9.8)(6.37 \times 10^6)}

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

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

From Moon:

Surface gravity on Moon: gM=GMMRM2=(6.67ร—10โˆ’11)(7.35ร—1022)(1.74ร—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=1.62ย m/s2g_M = 1.62 \text{ m/s}^2

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

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

vesc,Moon=2370ย m/s=2.37ย 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.