Gravitation AP/Honors Physics 1

Mr. Velazquez

Newton’s Law of Gravitation

• Newton was the first to make the connection between objects falling on Earth and the motion of the planets

• To illustrate this connection in Principia Mathematica Newton described the act of throwing a projectile horizontally from the top of a mountain

• Given a great enough velocity, the projectile would circle the Earth completely before returning to its starting point

Newton’s Law of Gravitation

𝐹𝐺 = 𝐺𝑚1𝑚2

𝑟2

𝑚1, 𝑚2 = point masses (kg)

𝑟 = distance between the centers of mass (m)

𝐺 = 6.67 × 10−11 N ∙ m2 kg2

𝒎𝟏

𝒎𝟐

𝑭𝑮

𝑭𝑮

𝒓

To describe how the gravitational force acts between two masses, Newton derived the following equation:

Newton’s Law of Gravitation a) Calculate the gravitational force between each of the

three pairs of masses below

b) Calculate the resultant gravitational force acting on mass 𝑚1

𝒎𝟏

𝒎𝟐

𝒎𝟑

22.5 m

12

.4 m

425 kg 152 kg

368 kg

Gravitational Acceleration for Spherical Bodies

𝑎 = 𝐺𝑀𝑃

𝑅𝑃2

𝑎 = acceleration due to gravity

𝑀𝑃 = mass of the sphere or planet

𝑅𝑃 = radius of the sphere or planet

𝐺 = 6.67 × 10−11 N ∙ m2 kg2

Earth’s Moon radius = 1,738 km

mass = 7.35 × 1022 kg

Mars radius = 3,389 km

mass = 6.39 × 1023 kg

Jupiter radius = 69,912 km

mass = 1.90 × 1027 kg

Uranus radius = 25,362 km

mass = 8.68 × 1025 kg

EXTRA CREDIT: Use the formula above to calculate the acceleration

due to gravity for the following bodies

Exit Ticket, Part 1 You will be estimating the gravitational force of a fist bump. First, choose a partner. Then, stand face-to-face and fist bump. Use a meter-stick to measure the distance from navel to navel (add an extra 8 cm for each male and 7 cm for each female in your pair). Then, convert each of your weights from pounds to kilograms (1 lb = 0.4536 kg). Once this is done, you can use Newton’s gravitation formula:

𝐹𝐺 = 𝐺𝑚1𝑚2

𝑟2

Gravitational Potential Energy

For objects at the surface of the Earth (where g is constant), potential energy can be calculated using 𝑈 = 𝑚𝑔ℎ. But as our distance from the surface increase, the value of g decreases. In addition, we want the potential energy from gravitation to become smaller the closer we get to Earth’s surface. It can be shown that the gravitational potential energy of a system where a mass 𝑚1 is a distance r from the center of the Earth (𝑚2) is:

𝑈 = −𝐺𝑚1𝑚2

𝑟

This potential energy is zero when the masses 𝑚1 and 𝑚2 are infinitely separated from each other.

Conservation of Energy By applying the laws of Conservation of Mechanical Energy, we can describe the total energy of an object with mass m, speed v, and a distance r away from the center of the Earth in the following way:

𝐸 = 𝐾 + 𝑈

𝐸 =1

2𝑚𝑣2 −

𝐺𝑚𝑀𝐸

𝑟

These energy equations can be used to calculate the speed and trajectory of any moving objects in space, from asteroids to comets to satellites.

Escape and Orbit Velocities

Suppose you wanted to send an object into space with enough velocity that it would eventually escape Earth’s gravitational pull—this refers to escape velocity. Or perhaps you wanted to send the object into a perfectly circular orbit around the Earth—or orbital velocity. These refer to specific magnitudes for the speed of the object, and we can compute them in different ways.

Escape and Orbit Velocities

The speed of the object will affect the type of motion—or rather, which conic section can be used to represent the motion.

Escape and Orbit Velocities

Escape and Orbital Velocities

For an object to have a perfectly circular orbit around Earth, the gravitational force must be exactly equal to the centripetal force. We use this fact to calculate orbital speed (𝑣𝑜 or 𝑣𝑐)

𝑚𝑣𝑜2

𝑟= 𝐺

𝑚𝑀𝐸

𝑟2

𝑣𝑜2 = 𝐺

𝑀𝐸

𝑟

𝑣𝑜 =𝐺𝑀𝐸

𝑟≈ 7,909 𝑚 𝑠

Gravitational Force

Centripetal Force

Orbital Velocity

Escape and Orbital Velocities For an object to leave Earth’s gravitational field, it must first achieve escape velocity. We calculate this by assuming the object will start with both kinetic and gravitational energy, and end up with neither.

𝐸1 = 𝐸2 𝐾 + 𝑈𝐺 = 0

1

2𝑚𝑜𝑣𝑒

2 −𝐺𝑚𝑜𝑀𝐸

𝑟= 0

𝑣𝑒2 = 2

𝐺𝑀𝐸

𝑟

𝑣𝑒 = 2𝐺𝑀𝐸

𝑟≈ 11,185 m s

Escape Velocity

Escape and Orbital Velocities

The only difference between escape velocity and orbital velocity is the square root of 2.

𝑣𝑒 = 𝑣𝑜 2

Exit Ticket, Part 2: Escape and Orbital Velocity

Find the escape and orbital velocity on Earth’s moon.

radius = 1,738 km

mass = 7.35 × 1022 kg

𝑣𝑒 = 2𝐺𝑀𝑚

𝑟 𝑣𝑜 =

𝐺𝑀𝑚

𝑟

Problem Set: Gravitation Pg. 378-379, #3, 6, 9, 12, 15, 18, 36, 39, 42, 45

Kepler’s Laws of Orbital Motion

Kepler’s First Law

Planets follow elliptical orbits, with the Sun at one focus of the ellipse.

Drawing Ellipses

Kepler’s Laws of Orbital Motion

Kepler’s Second Law

As a planet moves in its orbit, it sweeps out an equal amount of area in an equal amount of time.

Kepler’s Laws of Orbital Motion

Kepler’s Third Law

The square of the period, 𝑇2, of any planet is proportional to the cube of the semimajor axis (mean distance), 𝑟3, of its orbit. Or:

𝑇2 ∝ 𝑟3

𝑇2 = constant 𝑟3

𝑇2 =4𝜋2

𝐺𝑀𝑟3

Read your book (Chapter 12) for proof

of this constant