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
where N·m²/kg² is gravitational constant.
Vector form:
Gravitational Potential Energy
Choosing at :
Work to bring masses from infinity to separation :
Gravitational Field
Gravitational potential:
Escape Velocity
Minimum velocity to escape gravitational field:
Set total energy = 0:
For Earth: km/s
Circular Orbits
For circular orbit of radius :
Centripetal force = Gravitational force:
Orbital velocity:
Orbital period:
Orbital energy:
(Negative: bound orbit)
Note: (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:
Third Law (Harmonic Law):
where is semi-major axis.
For circular orbit ():
Elliptical Orbits
Energy:
where is semi-major axis.
Semi-major axis from energy:
Angular momentum:
where is eccentricity.
Eccentricity:
Perihelion and Aphelion
Perihelion (closest):
Aphelion (farthest):
Velocities:
(From energy and angular momentum conservation)
Hohmann Transfer Orbit
Efficient orbit change between circular orbits:
Transfer orbit energy:
Velocity changes:
Reduced Mass Problem
Two-body problem reduces to one-body with reduced mass:
Both orbit common center of mass.
Gravitational Force Inside Sphere
For uniform sphere, only mass at contributes:
(Linear with , like spring force!)
At center:
Shell Theorem
- Uniform spherical shell exerts no force on particle inside
- 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:
Numerator:
Denominator:
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: m
(a) Orbital speed:
For circular orbit:
Using → :
(b) Orbital period:
(c) Acceleration:
Centripetal acceleration:
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:
At infinity (just escaping):
Energy conservation:
Using :
From Earth:
From Moon:
Surface gravity on Moon:
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.
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