SAT Physics Gravity - Universal Gravity

SAT Physics Gravity - Universal Gravity

UNIVERSAL GRAVITY
Isaac Newton determined that the force of gravitational attraction, F_g, between two masses was directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Newton’s universal law of gravitation can be expressed as an equation

The letter G represents the universal gravitational constant, G= 6.67 x 10^-11 m³/kg • s².

Inverse Square Law
As the distance between two masses increases, the force of gravity between them decreases by the square of that distance. This means that a doubling of distance would result in quarter­ing of the gravitational force between the masses.

Calculating the Change in the Force of Gravity

Calculate the resulting force of gravitational attraction between two masses if one of the masses was to double and the distance between them was to triple.
 
WHAT'S THE TRICK?

The original force of attraction can be calculated using Newton’s law of gravity.

Substituting 2m, for m, and (3r)² for r² represents doubling of the mass and tripling of the distance, respectively.

Note that the quantity 3r is squared.

The force of gravity between the masses will be two-ninths of its original value.

GRAVITATIONAL FIELD
The influence, or alteration, of space surrounding a mass is known as a gravitational field. The gravitational field of Earth can be visually represented as several vector arrows pointing toward the surface of Earth, labeled g, as shown in Figure 9.1. The surface gravity field for Earth, g, is 10 meters per second squared. When an object of mass m is placed in Earth’s gravity field, it experiences a force of gravity, F_g , in the same direction as the gravity field.

Figure 9.1. The gravitation field of Earth

The effect on mass m in the gravity field can be solved as a force problem.

ma = mg
a = g

 It now becomes apparent that the acceleration of a mass in a gravity field equals the magnitude of the gravity field. Although there is a conceptual difference between the gravity field and the acceleration of gravity, they both have the same magnitude and direction.

Finding Surface Gravity
The gravity field of any planet is a function of the mass of the planet and the distance of the planet’s surface from its center. The exact relationship can be derived using two formulas for the force of gravity. When a mass m rests on the surface of Earth, the force of gravity can be determined using either the weight formula or Newton’s law of gravitation.

When these formulas are set equal to one another, mass m cancels. This results in a formula that describes both the strength of the gravity field and the acceleration due to gravity. Note that the mass of Earth is 5.98 X 10²⁴ kg and that the radius is 6.37 X 10⁶ m.

The gravity field near the surface of Earth is considered uniform. It has the same value at all points close to the surface of Earth. The magnitude of mass m placed on the surface of Earth does not matter since it cancels. Elephants and feathers are both in the same gravi­tational field and experience the same acceleration of 9.8 m/s². For the SAT Subject Test in Physics, the value for g will be rounded to 10 m/s² in order to make calculations easier.
Although the formula for g was derived on the surface of Earth, it can be generalized to solve for gravity on any planet or at a point in space near any planet.

In its generalized form, M is the mass of the planet and r is the distance measured from the center of the planet to the location where gravity, g, is to be calculated

Calculating the Surface Gravity of the Moon

Calculate the surface gravity of the Moon. The Moon has a mass of 7.36 x 10²² kilograms and a radius of 1.74 x 10⁶ meters.
 
WHAT'S THE TRICK?

Use the formula for finding the gravity of a planet and substitute in the values for the Moon.

This value is roughly one-sixth of the value of the surface gravity of Earth.
The SAT Subject Test in Physics does not allow you to use a calculator, and it is unlikely to present a problem requiring calculations that will be this involved. This example serves to demonstrate the universality of the equation for finding the gravitational field and the acceleration of gravity for any celestial object.

CIRCULAR ORBITS 
When a ball is thrown horizontally on Earth, it will follow a parabolic path toward the ground. During its flight, the ball simultaneously experiences a constant downward acceleration and a constant forward velocity. To an observer, it appears that the ball is moving in a parabola relative to a flat Earth.
Newton hypothesized that if a ball could be thrown with sufficient forward velocity, it would travel so quickly that the acceleration pulling it downward would not bring it to Earth. This is because the spherical Earth would curve out of the ball’s way as it fell. Today, satellites in orbit are able to do this with speeds exceeding 7,900 meters per second (17,500 miles per hour).
As discussed in Chapter 6 “Circular Motion,” objects experiencing an acceleration per­pendicular to their motion will move in a circle. The magnitude of their acceleration remains constant. However, their direction is constantly changing so that it always points toward the center of rotation.

Calculating Tangential Orbital Velocities

Figure 9.2. Satellite of mass m orbiting central body M

Equations for gravity and circular motion can be combined to determine the velocity of a satellite in a circular orbit. In Chapter 7, the following circular motion equations for centripetal acceleration and centripetal force were introduced.

Equations for gravity and circular motion can be combined to determine the velocity of a satellite in a circular orbit. In Chapter 7, the following circular motion equations for centripetal acceleration and centripetal force were introduced.As seen in Figure 9.2, the acceleration of gravity and the force of gravity are always directed toward the center of the circular path followed by orbiting mass m. They are the centripetal acceleration and force on satellites in circular orbits.

Rearranging and simplifying two of these equations will solve for orbital speed v.

In the above equation, mass M is the mass of a planet or star at the center of an orbiting satellite. Note that the formula does not contain m, the mass of the orbiting satellite itself. The speed of a satellite does not depend on its own mass. The satellite’s speed depends on the mass of the larger body it is orbiting and on the distance, r, from the center of the larger body. There apparently is an inverse relationship between speed and distance. Satellites closer to the central body orbit at higher speeds.

Calculating a Satellite’s Orbital Velocity

Calculate the speed of an orbiting satellite around Mars at a height of 3.57 x 10⁶ meters above the center of the planet. The mass of Mars is 6.42 x 10²³ kilograms.
 
WHAT'S THE TRICK?

Use the general equation for finding the velocity of a satellite above a planet of known mass, M, at radius r above the planet’s center.

As with Example 9.2, you should note that the mathematical complexity of this example will most likely not be on the SAT Subject Exam in Physics. Rather, you will more likely be tested on your understanding of the equation and how the orbital speed will be affected by changing the variables. Look for such examples in the questions at the end of this chapter and on the practice exams.
Understanding the formulas associated with orbital motion will help you answer many conceptual questions concerning changing variables. Consider the key formulas discussed thus far:

The SAT Subject Test in Physics will most likely ask questions based upon altering one or more of these variables and how that would affect orbital speed or the force of gravity. Actual numerical values will probably not be required since calculators are not allowed during the examination. For that reason, you should commit these fundamental gravity and orbital for­mulas to memory.

Determining the Effect of Changing Radius on an Orbiting Satellite

A satellite orbiting at a speed of v and a radius of r above the center of a planet climbs to a radius 2r. What is the satellite’s new orbital speed?
 
WHAT'S THE TRICK?

Orbital speed is determined by the following formula.

Substituting 2r for r represents doubling the orbital distance when the satellite moves from r to 2r. The right side of the equation shown below is now multiplied by √½. In order to maintain the equality, the left side of the equation needs to be multiplied by the same factor.

Doubling the orbital radius decreases the orbital velocity to √½ v.

KEPLER’S LAWS
Seventeenth-century astronomer Johannes Kepler deduced three laws describing the motion of planets around the Sun.
1. Planets orbit the Sim along an elliptical path where the Sun is at one of the two foci of the ellipse, 
    as shown in Figure 9.3.
2. A line drawn from the Sun to a planet would sweep out an equal area during an equal interval of 
    time. In Figure 9.3, area₁ = area₂

Figure 9.3 Kepler's law

3. The square of the period of a planet’s orbit is proportional to the cube of its radius in its orbit about
    the Sun.
These laws were deduced from Kepler’s analysis of observations made by Tycho Brahe. They were considered controversial by several of Kepler’s contemporary astronomers. The first law was particularly controversial. Most believed that planets orbited in perfect circles around the Sun.
Kepler’s second law illustrates that when a planet is closer to the Sun (known as perihe­lion), it will move at a faster orbital speed than when it is at its farthest point from the Sun (known as aphelion). This will allow the imaginary line between Sun and planet to sweep out an equal area during an equal interval of time.
The third law relates a mathematical relationship between orbital period and orbital radius.

T² ∝ T³

Committing this relationship to memory will help you answer conceptual questions regarding the effect on the period by changing the radius. This relationship works for all objects orbiting a common central mass.