12) if the mass of the earth and object is...
g. increased by a factor of 2, the $f_g$ is ____________ by a factor of ____________
h. increased by a factor of 3, the $f_g$ is ____________ by a factor of ____________
i. reduced by half, the $f_g$ is ____________ by a factor of ____________
13) at what point is the comet going to experience the greatest gravitational force and tangential velocity? why?
(image of a comets orbit with a sun and labeled points)
14) which of the following graphs accurately shows the force of gravity vs mass?
(four graphs labeled a, b, c, d with force (n) vs mass (kg))
15) which of the following graphs accurately shows the force of gravity vs distance from the center?
(four graphs labeled a, b, c, d with force (n) vs distance)
16) two students are sitting 1.5 m apart. one student has a mass of 70 kg and the other has a mass of 52 kg. what is the gravitational force between them (dont forget about g!!!!)?
17) what gravitational force does the moon produce on the earth if their centers are $3.84 \times 10^8$ m apart and the moon has a mass of $7.36 \times 10^{22}$ kg? (hint: the mass of the earth is on your formula chart.)

Question

12) if the mass of the earth and object is...
    g. increased by a factor of 2, the $f_g$ is ____________ by a factor of ____________
    h. increased by a factor of 3, the $f_g$ is ____________ by a factor of ____________
    i. reduced by half, the $f_g$ is ____________ by a factor of ____________
13) at what point is the comet going to experience the greatest gravitational force and tangential velocity? why?
    (image of a comets orbit with a sun and labeled points)
14) which of the following graphs accurately shows the force of gravity vs mass?
    (four graphs labeled a, b, c, d with force (n) vs mass (kg))
15) which of the following graphs accurately shows the force of gravity vs distance from the center?
    (four graphs labeled a, b, c, d with force (n) vs distance)
16) two students are sitting 1.5 m apart. one student has a mass of 70 kg and the other has a mass of 52 kg. what is the gravitational force between them (dont forget about g!!!!)?
17) what gravitational force does the moon produce on the earth if their centers are $3.84 \times 10^8$ m apart and the moon has a mass of $7.36 \times 10^{22}$ kg? (hint: the mass of the earth is on your formula chart.)

Accepted Answer

### Question 14
# Explanation:
## Step 1: Recall the Gravitational Force Formula
The gravitational force $ F_g $ between two objects is given by Newton's law of universal gravitation: $ F_g = G \frac{m_1 m_2}{r^2} $, where $ G $ is the gravitational constant, $ m_1 $ and $ m_2 $ are the masses of the two objects, and $ r $ is the distance between their centers. When considering the force of gravity on an object (e.g., an object near the Earth's surface), we can also write $ F_g = mg $, where $ g $ is the acceleration due to gravity (a constant near the Earth's surface). In the context of a graph of $ F_g $ vs. mass ($ m $) of an object, if we consider the force on the object due to the Earth, $ F_g = mg $, which is a linear relationship (since $ g $ is constant). So $ F_g $ is directly proportional to $ m $, meaning the graph should be a straight line with a positive slope.

## Step 2: Analyze the Graphs
- Graph A: Shows a straight line with a positive slope, indicating a direct proportionality ( $ F_g \propto m $ ), which matches $ F_g = mg $.
- Graph B: Shows a straight line with a negative slope, which would imply $ F_g $ decreases as $ m $ increases, contradicting $ F_g = mg $.
- Graph C: Shows a curve that decreases as $ m $ increases, which is not a linear relationship and does not match $ F_g = mg $.
- Graph D: Shows a curve that decreases as $ m $ increases, also not matching the linear relationship of $ F_g = mg $.

# Answer:
A. The graph with a positive linear slope (Graph A)

### Question 15
# Explanation:
## Step 1: Recall the Gravitational Force Formula for Distance
From Newton's law of universal gravitation, $ F_g = G \frac{m_1 m_2}{r^2} $. This shows that the gravitational force $ F_g $ is inversely proportional to the square of the distance $ r $ between the centers of the two objects ( $ F_g \propto \frac{1}{r^2} $ ). So as $ r $ increases, $ F_g $ decreases, and the relationship is a hyperbolic curve (not a straight line) because it's an inverse square relationship.

## Step 2: Analyze the Graphs
- Graph A: Straight line with positive slope, implying $ F_g \propto r $, which contradicts $ F_g \propto \frac{1}{r^2} $.
- Graph B: Straight line with negative slope, implying $ F_g \propto -r $, also contradicting the inverse square relationship.
- Graph C: Curve that decreases as $ r $ increases, but it's a different shape than an inverse square curve. Wait, no—wait, let's re-examine. Wait, the inverse square relationship $ y = \frac{k}{x^2} $ (where $ y = F_g $ and $ x = r $) is a curve that starts high when $ x $ is small and decreases rapidly at first, then more slowly as $ x $ increases. Wait, maybe I mislabeled. Wait, let's check the axes: Force (N) vs. Distance (m). So $ F_g = G \frac{m_1 m_2}{r^2} $, so $ F_g $ is inversely proportional to $ r^2 $. So the graph of $ F_g $ vs. $ r $ should be a hyperbola (a curve) where $ F_g $ decreases as $ r $ increases, and the rate of decrease slows as $ r $ gets larger (since it's $ 1/r^2 $, not $ 1/r $).

Wait, let's re-analyze the graphs:

- Graph A: Linear positive slope: $ F_g \propto r $ → wrong.
- Graph B: Linear negative slope: $ F_g \propto -r $ → wrong.
- Graph C: Let's see the shape. If it's a curve that decreases steeply at first then less steeply, but wait, $ 1/r^2 $ decreases more steeply as $ r $ increases? Wait, no: when $ r $ is small, $ 1/r^2 $ is large, and as $ r $ increases, $ 1/r^2 $ decreases, and the rate of decrease (the slope of the curve) becomes less steep (since the derivative of $ 1/r^2 $ is $ -2/r^3 $, which becomes less negative as $ r $ increases). Wait, maybe the graphs are labeled differently. Wait, the problem is "force of gravity vs distance from the center". So $ F_g = G \frac{Mm}{r^2} $, so $ F_g $ is inversely proportional to $ r^2 $. So the graph should be a curve where $ F_g $ decreases as $ r $ increases, and the curve is a hyperbola (like $ y = 1/x^2 $). Let's check the options:

- Graph D: Let's assume Graph D is the curve that represents $ y = 1/x^2 $ (starts high, decreases, and the curve gets flatter as $ r $ increases). Wait, maybe I mixed up C and D. Wait, the original problem's graphs for Q15: let's re-express. If $ F_g \propto 1/r^2 $, then as $ r $ increases, $ F_g $ decreases, and the relationship is a curve with a decreasing slope (getting less steep as $ r $ increases). So the correct graph should be the one that shows $ F_g $ decreasing as $ r $ increases, following an inverse square relationship. Among the options, the graph that is a curve (not linear) and shows $ F_g $ decreasing as $ r $ increases, with the curve's slope becoming less steep (i.e., the curve is concave up? Wait, $ y = 1/x^2 $ is concave up for $ x > 0 $). So if Graph D is the curve that looks like $ y = 1/x^2 $ (starts high, decreases, and the curve is concave up), that would be the correct one. Wait, maybe I made a mistake earlier. Let's re-express:

From $ F_g = G \frac{Mm}{r^2} $, $ F_g $ is inversely proportional to $ r^2 $, so the graph of $ F_g $ vs. $ r $ is a hyperbola (a curve) where $ F_g $ decreases as $ r $ increases, and the rate of decrease is proportional to $ 1/r^3 $ (the derivative), so the curve becomes less steep as $ r $ increases. So among the four graphs (A: linear positive, B: linear negative, C: curve decreasing steeply then less? Wait, no—maybe the labels are different. Wait, the problem's Q15 graphs: let's assume:

- Graph A: linear positive ( $ F_g $ increases with $ r $ ) → wrong.
- Graph B: linear negative ( $ F_g $ decreases with $ r $ linearly) → wrong, because it's inverse square, not inverse linear.
- Graph C: curve that decreases, but maybe inverse linear ( $ y = k/r $ ), which is a different curve (more like a hyperbola but with a different exponent).
- Graph D: curve that decreases, following $ y = k/r^2 $ (inverse square), which is the correct relationship.

So the correct graph is D.

# Answer:
D. The graph with the inverse square curve (Graph D)

### Question 16
# Explanation:
## Step 1: Identify the Formula and Values
We use Newton's law of universal gravitation: $ F_g = G \frac{m_1 m_2}{r^2} $, where:
- $ G = 6.674 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2 $ (gravitational constant)
- $ m_1 = 70 \, \text{kg} $ (mass of first student)
- $ m_2 = 52 \, \text{kg} $ (mass of second student)
- $ r = 1.5 \, \text{m} $ (distance between them)

## Step 2: Plug in the Values
First, calculate the product of the masses: $ m_1 m_2 = 70 \times 52 = 3640 \, \text{kg}^2 $

Then, calculate $ r^2 = (1.5)^2 = 2.25 \, \text{m}^2 $

Now, plug into the formula:

$ F_g = (6.674 \times 10^{-11}) \times \frac{3640}{2.25} $

First, calculate $ \frac{3640}{2.25} \approx 1617.78 $

Then, multiply by $ 6.674 \times 10^{-11} $:

$ F_g \approx 6.674 \times 10^{-11} \times 1617.78 \approx 6.674 \times 1617.78 \times 10^{-11} $

Calculate $ 6.674 \times 1617.78 \approx 6.674 \times 1600 + 6.674 \times 17.78 \approx 10678.4 + 118.7 \approx 10797.1 $

Then, $ 10797.1 \times 10^{-11} \approx 1.08 \times 10^{-7} \, \text{N} $ (approximate value, more precise calculation: $ 6.674e-11 * 3640 / 2.25 = (6.674 * 3640) / (2.25 * 1e11) = (24293.36) / (2.25e11) ≈ 1.08e-7 \, \text{N} $)

# Answer:
The gravitational force between them is approximately $ 1.08 \times 10^{-7} \, \text{N} $ (or more precisely, using $ G = 6.67 \times 10^{-11} $: $ F_g = 6.67e-11 * (70*52)/(1.5^2) = 6.67e-11 * 3640 / 2.25 ≈ 6.67e-11 * 1617.78 ≈ 1.079e-7 \, \text{N} $, so approximately $ 1.08 \times 10^{-7} \, \text{N} $)

### Question 17
# Explanation:
## Step 1: Identify the Formula and Values
We use Newton's law of universal gravitation: $ F_g = G \frac{m_1 m_2}{r^2} $, where:
- $ G = 6.674 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2 $
- $ m_1 = 5.972 \times 10^{24} \, \text{kg} $ (mass of Earth, from formula chart)
- $ m_2 = 7.36 \times 10^{22} \, \text{kg} $ (mass of Moon)
- $ r = 3.84 \times 10^8 \, \text{m} $ (distance between Earth and Moon's centers)

## Step 2: Plug in the Values
First, calculate $ m_1 m_2 = (5.972 \times 10^{24}) \times (7.36 \times 10^{22}) = 5.972 \times 7.36 \times 10^{24+22} \approx 43.95 \times 10^{46} = 4.395 \times 10^{47} \, \text{kg}^2 $

Then, calculate $ r^2 = (3.84 \times 10^8)^2 = 14.7456 \times 10^{16} = 1.47456 \times 10^{17} \, \text{m}^2 $

Now, plug into the formula:

$ F_g = (6.674 \times 10^{-11}) \times \frac{4.395 \times 10^{47}}{1.47456 \times 10^{17}} $

First, simplify the fraction: $ \frac{4.395 \times 10^{47}}{1.47456 \times 10^{17}} = \frac{4.395}{1.47456} \times 10^{47-17} \approx 2.98 \times 10^{30} $

Then, multiply by $ 6.674 \times 10^{-11} $:

$ F_g \approx 6.674 \times 10^{-11} \times 2.98 \times 10^{30} \approx 6.674 \times 2.98 \times 10^{19} \approx 19.89 \times 10^{19} = 1.99 \times 10^{20} \, \text{N} $ (approximate value; more precise calculation: $ 6.674e-11 * (5.972e24 * 7.36e22) / (3.84e8)^2 $)

Let's do it more precisely:

$ m_1 m_2 = 5.972e24 * 7.36e22 = 5.972 * 7.36 * 1e46 = 43.95392 * 1e46 = 4.395392e47 $

$ r^2 = (3.84e8)^2 = 14.7456e16 = 1.47456e17 $

$ F_g = 6.674e-11 * (4.395392e47 / 1.47456e17) = 6.674e-11 * (4.395392 / 1.47456) * 1e30 $

$ 4.395392 / 1.47456 ≈ 2.9807 $

$ 6.674e-11 * 2.9807e30 = 6.674 * 2.9807 * 1e19 ≈ 19.89 * 1e19 = 1.989e20 \, \text{N} $, so approximately $ 2.0 \times 10^{20} \, \text{N} $ (or more accurately, around $ 1.99 \times 10^{20} \, \text{N} $)

# Answer:
The gravitational force the Moon produces on the Earth is approximately $ 2.0 \times 10^{20} \, \text{N} $ (or more precisely, $ \approx 1.99 \times 10^{20} \, \text{N} $)

### Question 13
# Brief Explanations:
The comet experiences the greatest gravitational force and tangential velocity at the point closest to the Sun (the perihelion, which is the point labeled "A" in the diagram, assuming "A" is the closest point to the Sun).

### Reason for Gravitational Force:
According to Newton's law of universal gravitation, $ F_g = G \frac{Mm}{r^2} $, where $ M $ is the Sun's mass, $ m $ is the comet's mass, and $ r $ is the distance between the comet and the Sun. Gravitational force is inversely proportional to the square of the distance ($ F_g \propto \frac{1}{r^2} $). So when the comet is closest to the Sun (smallest $ r $), the gravitational force is the greatest.

### Reason for Tangential Velocity:
By the law of conservation of angular momentum, $ L = mvr $ (where $ L $ is angular momentum, $ v $ is tangential velocity, and $ r $ is the distance from the Sun). Angular momentum is conserved (assuming no external torques), so $ mvr = \text{constant} $. When $ r $ is smallest (closest to the Sun), $ v $ must be largest to keep $ L $ constant. Thus, tangential velocity is greatest at the closest point.

# Answer:
The comet experiences the greatest gravitational force and tangential velocity at the point closest to the Sun (e.g., point "A" in the diagram). The gravitational force is greatest there because $ F_g \propto \frac{1}{r^2} $ (smallest $ r $ gives largest $ F_g $). Tangential velocity is

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QUESTION IMAGE

Question

Explanation:

Question 14

Step 1: Recall the Gravitational Force Formula

Step 2: Analyze the Graphs

Step 1: Recall the Gravitational Force Formula for Distance

Step 2: Analyze the Graphs

Step 1: Identify the Formula and Values

Step 2: Plug in the Values

Answer:

Question 15