Solved Universal Law of Gravitation Worksheet Newtons | Chegg.com - Free Printable

Let's solve both problems step by step using Newton’s Universal Law of Gravitation:

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

Where:
- $ F $ is the gravitational force between two masses,
- $ G = 6.67 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2 $ (universal gravitational constant),
- $ m_1 $ and $ m_2 $ are the masses,
- $ r $ is the distance between the centers of the two masses.

---

Problem 1: Force of gravity on a space shuttle

Given:
- Mass of space shuttle, $ m_1 = 1.0 \times 10^5 \, \text{kg} $
- Altitude = 200.0 km = $ 2.00 \times 10^5 \, \text{m} $
- Radius of Earth ≈ $ 6.37 \times 10^6 \, \text{m} $
- So, distance from center of Earth to shuttle:
$ r = R_{\text{Earth}} + \text{altitude} = 6.37 \times 10^6 + 2.00 \times 10^5 = 6.57 \times 10^6 \, \text{m} $
- Mass of Earth, $ m_2 = 5.97 \times 10^{24} \, \text{kg} $ (standard value)
- $ G = 6.67 \times 10^{-11} $

Now calculate:

$$
F = G \frac{m_1 m_2}{r^2}
= (6.67 \times 10^{-11}) \frac{(1.0 \times 10^5)(5.97 \times 10^{24})}{(6.57 \times 10^6)^2}
$$

Step-by-step:

1. Numerator:
$ (1.0 \times 10^5) \times (5.97 \times 10^{24}) = 5.97 \times 10^{29} $

2. Denominator:
$ (6.57 \times 10^6)^2 = 43.1649 \times 10^{12} = 4.31649 \times 10^{13} $

3. Fraction:
$ \frac{5.97 \times 10^{29}}{4.31649 \times 10^{13}} \approx 1.383 \times 10^{16} $

4. Multiply by $ G $:
$ F = (6.67 \times 10^{-11}) \times (1.383 \times 10^{16}) \approx 9.23 \times 10^5 \, \text{N} $

✔ Answer for Problem 1:
The force of gravity on the space shuttle is approximately $ 9.23 \times 10^5 \, \text{N} $ (about 923,000 newtons).

> Note: This is less than the weight at Earth's surface ($ mg = 10^5 \times 9.8 = 9.8 \times 10^5 \, \text{N} $), which makes sense because it's at higher altitude.

---

Problem 2: Gravity between you and your car

Given:
- Your mass, $ m_1 = 55 \, \text{kg} $
- Car mass, $ m_2 = 1234 \, \text{kg} $
- Distance between centers, $ r = 3 \, \text{m} $
- $ G = 6.67 \times 10^{-11} $

$$
F = G \frac{m_1 m_2}{r^2}
= (6.67 \times 10^{-11}) \frac{(55)(1234)}{3^2}
$$

Step-by-step:

1. $ m_1 m_2 = 55 \times 1234 = 67,870 $

2. $ r^2 = 9 $

3. $ \frac{67,870}{9} \approx 7,541.11 $

4. $ F = (6.67 \times 10^{-11}) \times 7,541.11 \approx 5.03 \times 10^{-7} \, \text{N} $

✔ Answer for Problem 2:
The gravitational force between you and your car is about $ 5.03 \times 10^{-7} \, \text{N} $ — extremely small!

> This is why we don’t feel gravitational attraction between everyday objects — it's negligible compared to other forces like friction or normal force.

---

✔ Final Answers:

1. Force on space shuttle: $ \boxed{9.23 \times 10^5 \, \text{N}} $
2. Force between you and car: $ \boxed{5.03 \times 10^{-7} \, \text{N}} $

Let me know if you'd like these values rounded differently or want to compare them to Earth's surface gravity!