GRAVITY 1 - A worksheet with questions on gravitational force and Newton's law of gravitation.

It looks like you've uploaded a worksheet titled "GRAVITY I" with several problems related to gravitational force. Since I can't view images directly, I'll solve each problem based on the text you provided. Let's go through them one by one and explain the solutions.

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Problem 1:

Martha (50 kg) is attracted to Stewart (70 kg) who sits 4 m away. What is the gravitational attraction between them? G = 6.67 × 10⁻¹¹

We use Newton’s Law of Universal Gravitation:

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

Where:
- \( G = 6.67 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2 \)
- \( m_1 = 50 \, \text{kg} \)
- \( m_2 = 70 \, \text{kg} \)
- \( r = 4 \, \text{m} \)

\[
F = (6.67 \times 10^{-11}) \frac{(50)(70)}{4^2} = (6.67 \times 10^{-11}) \frac{3500}{16}
\]
\[
F = (6.67 \times 10^{-11}) \times 218.75 = 1.46 \times 10^{-8} \, \text{N}
\]

✔ Answer: \( 1.46 \times 10^{-8} \, \text{N} \)

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Problem 2:

Stewart (70 kg) is attracted to Ms. Littles (60 kg) who sits 2 m away. What is the gravitational attraction between them?

Same formula:

\[
F = (6.67 \times 10^{-11}) \frac{(70)(60)}{2^2} = (6.67 \times 10^{-11}) \frac{4200}{4}
\]
\[
F = (6.67 \times 10^{-11}) \times 1050 = 7.00 \times 10^{-8} \, \text{N}
\]

✔ Answer: \( 7.00 \times 10^{-8} \, \text{N} \)

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Problem 3:

Under the Big Top (trapeze), Ella (2500 kg) is attracted to Planet, the 5,000 kg elephant. They are separated by 8 m. What is the gravitational force between them?

Again:

\[
F = (6.67 \times 10^{-11}) \frac{(2500)(5000)}{8^2} = (6.67 \times 10^{-11}) \frac{12,500,000}{64}
\]
\[
F = (6.67 \times 10^{-11}) \times 195,312.5 = 1.30 \times 10^{-5} \, \text{N}
\]

✔ Answer: \( 1.30 \times 10^{-5} \, \text{N} \)

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Problem 4:

Mars is \( 3.85 \times 10^{23} \, \text{kg} \) and Earth is \( 5.98 \times 10^{24} \, \text{kg} \). The distance between them is \( 4.42 \times 10^7 \, \text{m} \). What is the gravitational attraction between them?

\[
F = (6.67 \times 10^{-11}) \frac{(3.85 \times 10^{23})(5.98 \times 10^{24})}{(4.42 \times 10^7)^2}
\]

First, compute numerator:
\[
(3.85 \times 10^{23})(5.98 \times 10^{24}) = 2.2613 \times 10^{48}
\]

Denominator:
\[
(4.42 \times 10^7)^2 = 1.9536 \times 10^{15}
\]

Now divide:
\[
\frac{2.2613 \times 10^{48}}{1.9536 \times 10^{15}} = 1.157 \times 10^{33}
\]

Multiply by G:
\[
F = (6.67 \times 10^{-11})(1.157 \times 10^{33}) = 7.72 \times 10^{22} \, \text{N}
\]

✔ Answer: \( 7.72 \times 10^{22} \, \text{N} \)

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Problem 5:

The Sun has a mass of \( 1.99 \times 10^{30} \, \text{kg} \) and is \( 1.5 \times 10^{11} \, \text{m} \) from Earth. The planet Earth is \( 5.98 \times 10^{24} \, \text{kg} \). What is the gravitational attraction between the sun and Earth?

\[
F = (6.67 \times 10^{-11}) \frac{(1.99 \times 10^{30})(5.98 \times 10^{24})}{(1.5 \times 10^{11})^2}
\]

Numerator:
\[
(1.99 \times 10^{30})(5.98 \times 10^{24}) = 1.19002 \times 10^{55}
\]

Denominator:
\[
(1.5 \times 10^{11})^2 = 2.25 \times 10^{22}
\]

Divide:
\[
\frac{1.19002 \times 10^{55}}{2.25 \times 10^{22}} = 5.288 \times 10^{32}
\]

Multiply by G:
\[
F = (6.67 \times 10^{-11})(5.288 \times 10^{32}) = 3.53 \times 10^{22} \, \text{N}
\]

✔ Answer: \( 3.53 \times 10^{22} \, \text{N} \)

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Problem 6:

If the moon were suddenly to grow twice as large in mass, how would that affect the force of gravitation with the Earth?

Gravitational force is proportional to the product of the masses:

\[
F \propto m_1 m_2
\]

If the Moon’s mass doubles (\( m_{\text{moon}} \to 2m_{\text{moon}} \)), then:

\[
F_{\text{new}} = G \frac{(2m_{\text{moon}}) m_{\text{earth}}}{r^2} = 2 \times F_{\text{original}}
\]

✔ Answer: The gravitational force would double.

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Problem 7:

Putting the moon back to its original size, how far would it have to move to halve the distance from the Earth, how would that affect the force of gravitation with the Earth?

Force depends inversely on the square of the distance:

\[
F \propto \frac{1}{r^2}
\]

If distance is halved, then:

\[
F_{\text{new}} = \frac{1}{(r/2)^2} = \frac{1}{r^2/4} = 4 \times F_{\text{original}}
\]

So the force becomes four times stronger.

✔ Answer: The gravitational force would quadruple (increase by a factor of 4).

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✔ Final Answers Summary:

1. \( 1.46 \times 10^{-8} \, \text{N} \)
2. \( 7.00 \times 10^{-8} \, \text{N} \)
3. \( 1.30 \times 10^{-5} \, \text{N} \)
4. \( 7.72 \times 10^{22} \, \text{N} \)
5. \( 3.53 \times 10^{22} \, \text{N} \)
6. The force would double
7. The force would quadruple

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