Study Guide: Circular Motion, Gravity, Work and Energy

Q1. What is the magnitude of the gravitational force acting on the earth due to the sun? Express your answer in newtons.

Background

Topic: Newton's Law of Universal Gravitation

This question tests your understanding of how to calculate the gravitational force between two massive objects (the earth and the sun) using Newton's law of universal gravitation.

Key formula:

$F = G \frac{m_1 m_2}{r^2}$

Where:

• $F$ = gravitational force (in newtons)

• $G$ = gravitational constant ($6.674 \times 10^{-11}$ N·m$^2$/kg$^2$)

• $m_1$ = mass of the earth ($5.98 \times 10^{24}$ kg)

• $m_2$ = mass of the sun ($1.99 \times 10^{30}$ kg)

• $r$ = distance between the earth and sun (must be in meters)

Step-by-Step Guidance

1. Identify the values for $m_1$, $m_2$, and $r$ from the problem statement. Make sure all values are in SI units (kg for mass, meters for distance).

2. Convert the distance between the earth and sun to meters if it is given in miles or kilometers. (1 mile = 1609 m, 1 km = 1000 m)

3. Write out the formula for gravitational force: $F = G \frac{m_1 m_2}{r^2}$

4. Plug the values for $G$, $m_1$, $m_2$, and $r$ into the formula, but do not calculate the final value yet.

Try solving on your own before revealing the answer!

Final Answer: $3.54 \times 10^{22}$ N

$F = (6.674 \times 10^{-11}) \frac{(5.98 \times 10^{24})(1.99 \times 10^{30})}{(1.496 \times 10^{11})^2}$

This calculation uses the correct SI units and values for the earth-sun system, resulting in a very large force due to their massive sizes and relatively close distance.

Q2. At the moment shown in the figure of the earth and sun, what is the direction of the gravitational force acting on the earth?

Background

Topic: Direction of Gravitational Force

This question tests your understanding of the vector nature of gravitational force and how it always acts along the line joining the centers of two masses.

Key concept:

• The gravitational force between two objects always points directly toward the other object.

Step-by-Step Guidance

1. Recall that gravitational force is attractive and acts along the line connecting the centers of mass of the two objects.

2. For the earth and sun, the force on the earth due to the sun points directly toward the sun.

Try solving on your own before revealing the answer!

Final Answer: The force always points directly toward the sun.

As the earth moves in its orbit, the direction of the gravitational force acting on it always points toward the sun.

Q3. What is the direction of the net gravitational force acting on a space probe halfway between the earth and the sun?

Background

Topic: Net Gravitational Force and Vector Addition

This question tests your ability to analyze the net force on an object due to multiple sources of gravity, considering both magnitude and direction.

Key formula:

$F = G \frac{m_1 m_2}{r^2}$

Key concept:

• Both the earth and sun exert gravitational forces on the probe. The net force is the vector sum of these forces.

Step-by-Step Guidance

1. Calculate the gravitational force exerted by the earth on the probe using the formula above, with $r$ as the distance from the probe to the earth.

2. Calculate the gravitational force exerted by the sun on the probe, with $r$ as the distance from the probe to the sun.

3. Compare the magnitudes of the two forces. Since the sun is much more massive, its force will be stronger.

4. Determine the direction of the net force by considering the vector sum: both forces act along the line connecting the earth and sun, but toward each respective body.

Try solving on your own before revealing the answer!

Final Answer: The net force points toward the sun.

Because the sun is much more massive than the earth, the probe experiences a stronger gravitational pull toward the sun at the midpoint.

Q4. What is the value of the composite constant $\left(\frac{m_e}{r_e^2}\right)$ for the earth, to be used in calculating gravitational acceleration?

Background

Topic: Gravitational Acceleration Near Earth's Surface

This question tests your understanding of how the mass and radius of the earth combine to determine the gravitational acceleration $g$ near the surface.

Key formula:

$g = G \frac{m_e}{r_e^2}$

Where:

• $g$ = gravitational acceleration near Earth's surface (in m/s$^2$)

• $G$ = gravitational constant

• $m_e$ = mass of the earth

• $r_e$ = radius of the earth

Step-by-Step Guidance

1. Write out the formula for gravitational acceleration: $g = G \frac{m_e}{r_e^2}$

2. Plug in the values for $m_e$ and $r_e$ (make sure $r_e$ is in meters).

3. Calculate the composite constant $\left(\frac{m_e}{r_e^2}\right)$, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer: $1.54 \times 10^{20}$ kg/m$^2$

This value is used in the calculation of $g$ and is derived from the mass and radius of the earth.

Q5. What is the radius $r$ of the turn if $\theta = 20.0^\circ$ (assuming the car moves in a circle)? Express your answer in meters.

Background

Topic: Circular Motion and Centripetal Force

This question tests your understanding of how to relate the angle of a turn to the radius of the circular path, using concepts from circular motion and Newton's laws.

Key formula:

$a_c = \frac{v^2}{r}$

Where:

• $a_c$ = centripetal acceleration

• $v$ = velocity of the car

• $r$ = radius of the turn

Step-by-Step Guidance

1. Identify the relationship between the angle $\theta$ and the radius $r$ for a car moving in a circle.

2. Apply Newton's second law to relate the net force to the centripetal acceleration.

3. Set up the equation for centripetal acceleration and solve for $r$ in terms of $v$ and $a_c$.

4. Plug in the given values, but do not compute the final radius yet.

Try solving on your own before revealing the answer!

Final Answer: $r = \frac{v^2}{a_c}$

By rearranging the centripetal acceleration formula, you can solve for the radius of the turn.