Gravitation — Kepler and Newton
Law of universal gravitation, gravitational field and potential, Kepler's laws, planetary orbits.
Complete Theory
4Newton's law states that every pair of point masses attracts each other with a force proportional to the product of their masses and inversely proportional to the square of the distance between them: .
Key properties:
Key properties:
- Always attractive: gravity never repels.
- Infinite range: the force never reaches exactly zero, no matter how far the bodies are.
- Inverse-square law: doubling the distance reduces the force by a factor of 4; tripling reduces it by 9. This is a geometric consequence of the force "spreading" over a sphere of area .
- Acts on point masses: for spherically symmetric bodies (like planets) the law holds exactly with measured between centres (Shell Theorem).
The gravitational field is defined as the gravitational force per unit mass at a point. It points toward the source mass and falls off as .
The gravitational potential is negative everywhere (by convention ). The potential energy of a mass in the field is . The field is the negative gradient of the potential: .
For a body in orbit around a central mass the total mechanical energy determines the orbit type:
The gravitational potential is negative everywhere (by convention ). The potential energy of a mass in the field is . The field is the negative gradient of the potential: .
For a body in orbit around a central mass the total mechanical energy determines the orbit type:
- : bound orbit — ellipse (circle is the special case ).
- : body just barely escapes — parabolic trajectory; this defines escape velocity.
- : body escapes with kinetic energy to spare — hyperbolic trajectory.
Kepler's three empirical laws, derived from precise astronomical observations, are exact consequences of Newton's law of gravitation:
First Law (Law of Orbits): every planet moves in an ellipse with the Sun at one focus. The shape is described by eccentricity : for a circle , for a highly elongated ellipse .
Second Law (Law of Areas): the radius vector from Sun to planet sweeps equal areas in equal times. This is a direct consequence of the conservation of angular momentum: because gravity is a central force (no torque about the Sun), is constant, hence . A planet moves fastest at perihelion and slowest at aphelion.
Third Law (Harmonic Law): the square of the orbital period is proportional to the cube of the semi-major axis: . The proportionality constant is the same for all bodies orbiting the same central mass, so comparing two planets gives .
First Law (Law of Orbits): every planet moves in an ellipse with the Sun at one focus. The shape is described by eccentricity : for a circle , for a highly elongated ellipse .
Second Law (Law of Areas): the radius vector from Sun to planet sweeps equal areas in equal times. This is a direct consequence of the conservation of angular momentum: because gravity is a central force (no torque about the Sun), is constant, hence . A planet moves fastest at perihelion and slowest at aphelion.
Third Law (Harmonic Law): the square of the orbital period is proportional to the cube of the semi-major axis: . The proportionality constant is the same for all bodies orbiting the same central mass, so comparing two planets gives .
Escape velocity is the minimum launch speed needed to escape a planet's gravity without further propulsion. It is derived by setting the total mechanical energy to zero (): , giving .
Note that escape velocity is independent of the mass of the escaping object and of the direction of launch — it only depends on the mass and radius of the planet.
For a circular orbit at radius , gravity provides the centripetal force: , giving . Comparing with : , so escape speed is always times the circular orbital speed at the same altitude.
The total energy of an elliptical orbit depends only on the semi-major axis : . A lower orbit has more negative energy (more tightly bound) and — counterintuitively — a higher speed, since .
Note that escape velocity is independent of the mass of the escaping object and of the direction of launch — it only depends on the mass and radius of the planet.
For a circular orbit at radius , gravity provides the centripetal force: , giving . Comparing with : , so escape speed is always times the circular orbital speed at the same altitude.
The total energy of an elliptical orbit depends only on the semi-major axis : . A lower orbit has more negative energy (more tightly bound) and — counterintuitively — a higher speed, since .
Worked Examples
2Example 1Escape velocity from Earth and Moon
Given
Earth mass
Earth radius
Moon mass
Moon radius
Find
from Earth's surface
from Moon's surface
Comparison between the two
Step-by-step solution
1Problem: compute the minimum speed needed to launch a rocket from the Earth's surface and from the Moon's surface so that it permanently escapes the gravitational pull (reaching infinity with zero speed).
2General formula: escape velocity follows from the energy balance: at infinity both and are zero (by convention), so at the surface we must have . Solving: . Note that does not depend on the mass of the launched body, only on the mass and radius of the celestial body.
3Escape velocity from Earth: (about 40284 km/h).
4Escape velocity from the Moon: (about 8532 km/h).
5Comparison: the lunar escape velocity is about 1/5 of Earth's. This is why the Apollo mission needed an enormous rocket (Saturn V) to leave Earth, but a small lunar module sufficed to depart from the Moon. On Jupiter () it would be extremely difficult; on a black hole, , so not even light can escape.
✓ Final result: ,
Example 2Geostationary orbit altitude
Given
required orbital period (24 hours)
Earth mass
Find
Orbital radius of a geostationary orbit
Altitude above Earth's surface
Orbital speed
Step-by-step solution
1Problem: a geostationary satellite remains above the same point on Earth's surface. Its orbital period must therefore equal Earth's rotation period: . We need to find the required orbital distance.
2Circular orbit condition: the gravitational force must equal the centripetal force: , with . This gives: .
3Radius calculation: . Taking the cube root: from Earth's centre.
4Altitude above surface: (about 35800 km). Geostationary satellites orbit at roughly 36000 km altitude.
5Orbital speed: (about 11050 km/h). For comparison, a low Earth orbit (LEO, 200 km altitude) has orbital speed of about 7.8 km/s and takes 90 minutes per revolution.
6Importance: geostationary satellites are used for telecommunications, satellite TV, and meteorology because their fixed position relative to Earth allows antennas to be pointed once and remain aimed.
✓ Final result: , ,
Exercises with Solutions
2Exercise 1Elliptical orbitHard
Problem to solve
A planet orbits the Sun on an elliptical orbit with semi-major axis (Astronomical Units, 1 AU = ) and eccentricity . Given: , . Determine: (a) the orbital period in years, (b) the speed at perihelion and aphelion , knowing that and .
Given data
a = 3.74\times10^{11}\,m (semi-major axis)e = 0.4 (eccentricity)GM_\odot = 1.327\times10^{20}\,m^3/s^2 (solar constant)
Step-by-step solution
1Setup: we use Kepler's third law for the period and conservation of angular momentum (Kepler's second law) for the velocities.
2(a) Orbital period: , hence . Convert to years: , so (3 years and about 11 months).
3(b) Speeds at perihelion and aphelion: the distances are and . From energy conservation: . From angular momentum conservation: (at perihelion and aphelion ).
4From Kepler's second law: . From the energy equation: → (about 28.8 km/s). Hence (about 12.3 km/s).
5Verification: Earth (, ) has and . Our planet is farther out () and much more eccentric, so the difference between perihelion and aphelion is enormous (28.8 vs 12.3 km/s).
✓ Final answer: , ,
Exercise 2Three collinear massesHard
Problem to solve
Three point masses are arranged along the x-axis: at , at , at . Determine the net gravitational force (magnitude and direction) acting on .
Given data
m_1 = 5\times10^{10}\,kg (at x=0)m_2 = 2\times10^{10}\,kg (at x=4\,m)m_3 = 3\times10^{10}\,kg (at x=10\,m)
Step-by-step solution
1Setup: the force on is the vector sum of the forces exerted by and . All forces lie on the x-axis, so we sum them algebraically (with sign).
2Force from on : toward (left), because attracts .
3Force from on : toward (right), because attracts .
4Net force: (toward the left, i.e. toward ). Mass attracts more strongly than , even though is more massive, because is closer.
5Interpretation: the gravitational force decays as , so proximity matters more than mass. is 4 m away, is 6 m away: the distance ratio is 1.5, the force ratio due to distance alone is , which outweighs the mass ratio (), giving .
✓ Final answer: toward (left)
Keep studying
Recommended Books
Introductory
Physics for Scientists and Engineers
Buy on Amazon →
Advanced
Classical Mechanics
Buy on Amazon →
As an Amazon Associate I earn from qualifying purchases.
🔥
Integrative Problems
Problems combining all chapters — exam levelProblem 1Tower, Ballistic Pendulum, and Keplerian OrbitEXTREME
A cannon is placed on top of a tower tall and fires a projectile of horizontally at .
The projectile strikes and embeds in a wooden block hanging from a rope of length (ballistic pendulum), at ground level.
The Earth-Moon system is then used as a reference for Kepler's third law.
The projectile strikes and embeds in a wooden block hanging from a rope of length (ballistic pendulum), at ground level.
The Earth-Moon system is then used as a reference for Kepler's third law.
📌 Problem data
(a)Uniformly Accelerated Motion(b)Inelastic Collision(c)Potential Energy + Pendulum(d)Moment of Inertia — Rigid Body(e)Gravitation — Kepler's Third Law
Problem 2Spring, Rolling Disk, Inclined Plane Collision, and ConservationEXTREME
A spring (, compressed ) launches a solid disk (, ) up an inclined plane (, , ) that rolls without slipping.
At the top the disk is launched horizontally and strikes a pendulum (, ) — perfectly inelastic collision. What is asked (solved below, a→e): (a) the disk's speed at the top of the plane; (b) the range and impact speed of the horizontal launch; (c) the speed after the inelastic collision with the pendulum and the energy lost; (d) the pendulum's maximum angle, the maximum tension, and whether it completes the loop; (e) the full energy balance (from spring to maximum angle).
At the top the disk is launched horizontally and strikes a pendulum (, ) — perfectly inelastic collision. What is asked (solved below, a→e): (a) the disk's speed at the top of the plane; (b) the range and impact speed of the horizontal launch; (c) the speed after the inelastic collision with the pendulum and the energy lost; (d) the pendulum's maximum angle, the maximum tension, and whether it completes the loop; (e) the full energy balance (from spring to maximum angle).
📌 Problem data
(a)Energy + Rigid Body (rolling)(b)Kinematics — Projectile(c)Inelastic Collision + CM(d)Pendulum Dynamics + Forces(e)Conservation Laws — Complete Energy Balance