### SAT Physics Gravity - Universal Gravity

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**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

^{-11}m

^{3}/kg • s

^{2}.

**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**

**WHAT'S THE TRICK?**

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

^{2}for r

^{2}represents doubling of the mass and tripling of the distance, respectively.

**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

**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.

^{24}kg and that the radius is 6.37 X 10

^{6}m.

^{2}. For the SAT Subject Test in Physics, the value for g will be rounded to 10 m/s

^{2}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.

**Calculating the Surface Gravity of the Moon**

^{22 }kilograms and a radius of 1.74 x 10

^{6}meters.

**WHAT'S THE TRICK?**

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

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.

**Calculating a Satellite’s Orbital Velocity**

^{6 }meters above the center of the planet. The mass of Mars is 6.42 x 10

^{23}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.

Understanding the formulas associated with orbital motion will help you answer many conceptual questions concerning changing variables. Consider the key formulas discussed thus far:

**Determining the Effect of Changing Radius on an Orbiting Satellite**

**WHAT'S THE TRICK?**

Orbital speed is determined by the following formula.

**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, area

_{1}= area

_{2}

**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.

T

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.^{2}∝ T^{3}
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