Summer 2012 PHYS 172: Modern Mechanics Lecture 5 – Gravity Read

The gravitational force law m1m1 m2m2 Newton m1m1 m2m2 Cavendish Gravitational constant

Features of gravitational force gravity is always attractive gravity is an inverse square law the force depends upon the product of the masses

Distance between two objects Point object: idealized object which has no size, all mass is in one point If distance between the two objects is >> than their size, can model the objects as point-masses Real objects: have size Special case: spherical objects (spherical symmetry) Uniform-density spheres interact gravitationally in exactly the same way as if all their mass were concentrated at the center of the sphere. Can model as a point mass!

Clicker question #1 What is the distance between these two spheres to be used in gravitational law? A B

Gravitational force on a planet starplanet 1. Calculate 2. Distance 3. Unit vector: 3. Force: directionmagnitude

Gravitational force on a planet starplanet Checking results: 1.Diagram 2.Order of magnitude 3.Units 4.Unit vector Clicker question # 3: What is the gravitational force exerted by the planet on the star? A) The same B) C)

Gravitational force near the Earth’s surface RERE m ~ The same for all objects on surface Gravitational field The magnitude: g = 9.8 N/kg M E = ×10 24 kg R E = 6.37 ×10 6 m

Reciprocity: Newton’s 3rd law Force magnitudes are the same Directions are opposite They act on different objects Reciprocity (Newton’s 3 rd law): The forces of two objects on each other are always equal and are directed in opposite directions NOTE: Velocity-dependent forces (e.g., magnetic forces) do not obey Newton’s 3 rd law!

The gravitational force exerted by a planet on one of its moons is 3e23 newtons when the moon is at a particular location. If the mass of the moon were three times as large, what would the force on the moon be? A) 1e23 N B) 3e23 N C) 6e23 N D) 9e23 N Q3.4.a

The gravitational force exerted by a planet on one of its moons is 3e23 newtons when the moon is at a particular location. If the distance between the moon and the planet was doubled, what would the force on the moon be? A) 6e23 N B) 3e23 N C) 1.5e23 N D) 0.75e23 N E) 0.33e23 N Q3.4.b

Fixed star position : m Initial planet position: m Calculate the vector that points from the star to the planet. A) m B) m C) m D) m E) We don’t have enough information to find the vector Q3.4.c:

The relative position vector from the star to the planet is: m. What is the distance between the star and the planet? A) 0.50e11 m B) 1.50e11 m C) 1.58e11 m D) 2.00e11 m E) 2.50e11 m Q3.4.d

Distance from star to planet: 1.58e11 m Star’s mass: 1e30 kg Planet’s mass: 5e24 kg G = 6.7e-11 N · m 2 /kg 2 Calculate the magnitude of the gravitational force that the star exerts on the planet. A) 1.34e-8 N B) 2.68e-2 N C) 1.34e22 N D) 2.12e33 N E) 5.3e55 N Q3.4.e:

Relative position vector from star to planet is m Distance from star to planet is 1.58e11 m Find the unit vector pointing from the star to the planet A) B) C) D) E) Q3.4.f:

Calculate the gravitational force exerted by the star on the planet (remember that force is a vector) A) N B) N C) N D) N E) N Q3.4.g: