1. Note: F = w = mg also, so g = Gm1/r2, acceleration due to gravity
Text : Griffith and Brosing
Basic Physics : PHYS:1400, Lecture 10, Feb , Wed
Circular Motion, the Planets and Gravity Ch 5
Newton’s law of Universal Gravitation
It states that the Gravitational force between two objects is proportional to their masses and inversely proportional to the square of the distance between the two objects.
G (the universal gravitational constant) = 6.67 × 1011 Nm2/kg2 .
Some of Kepler’s laws were explained by Newton assuming an inverse square law of gravitational attraction.
Note: F = w = mg also, so g = Gm1/r2, acceleration due to gravity
The gravitational force is attractive and acts along the line joining the center of the two masses.
also, F1 = F2 .

2. If lines are drawn radiating outward from a point mass, the areas intersected by these lines increase in proportion to r2.
The force exerted by the mass on a second mass might become weaker in proportion to 1/r2.
Three equal masses m1 = m2 = m3 are shown below, with the distances between them. What will be the direction of the net force acting on m2? And why ?
To the left.
To the right.
The forces cancel such that the total force is zero.
It is impossible to determine from the figure.

3. Phases of the moon result from the changes in the positions of the moon, Earth, and the sun.

4. Escape velocity of a gravitational field: artificial satellites and motion of a projectile:
Once a projectile is thrown, it goes as far as it can depending on its initial velocity, before falling to the surface of Earth. If the velocity is high enough, it never touches Earth’s surface, continuously “falling” but never arriving on the surface (because of roundness of Earth).

5. A car travels around a non-banked curve with constant speed
A car travels around a non-banked curve with constant speed. Sketch a diagram showing all of the forces acting on the car. What is the direction of the net force acting on the car?
The net force is toward the center of the curve, radially inward.
Two masses are separated by a distance r. If the distance between them is doubled, is the force of interaction doubled, halved, or changed by some other amount? Explain.
Since the gravitational force of attraction varies as 1/r2, doubling the distance between them will result in ¼ of the original force acting between them.

6. A synchronous satellite is one that does not move relative to the surface of the Earth; it is always above the same location. Why does such a satellite not just fall straight down to the Earth? Explain.
Just because it is always at the same point above the Earth, it is not standing “still”, it simply is moving at the same rate as the Earth, it has the same rotational period as the diurnal rotation of Earth. It is held in place by the centripetal force, its centripetal acceleration around the Earth.
How much larger is the required centripetal acceleration of a car rounding a curve at 60 MPH that of than of the one rounding the same curve at 30 MPH?
Magnitude of centripetal acceleration = v2/r
So, for car at 30 MPH, ac30 = V302/r and for car at 60 MPH ac60 = V602/r = (2V30)2/r = 4 V302/r = 4 ac30
So, four times as much.

7. One year of a planet X is three times as long as the Earth year; how far away from the sun is planet X compared to that of the Earth?
Use Kepler’s third law, T2  r3 , that is T2/r3 = constant for all planets around the sun
TE2/rE3 = TX2/rX3 ; so, rX3 = TX2/(TE2/rE3) = (TX2/TE2) × rE3 = 9 rE3 ; so , rX = 3 9rE3  2.1 rE
Planet X is almost 2.1 times the distance away from the sun that that of the Earth

8. Practice the examples and go over the topics done in class
Practice the examples and go over the topics done in class. For Friday, read Chapter 6, Sections 1, 2 and 3.
Attend TILE class and the discussion sections on Mon or Tues.
HW set 4 is uploaded, due before or by Monday, the usual time.