Captain Cutshaw is a 75.9 kg astronaut on board an experimental space station that orbits the Moon at an altitude of 269 km above the Moon's surface. Since the mass and average radius of the Moon are 7.350 x 1022 kg and 1740 km, respectively, how large must the gravitational force (in N) be that acts on Captain Cutshaw?
Captain Cutshaw is a 75.9 kg astronaut on board an experimental space station that orbits the Moon at an altitude of 269 km above the Moon's surface. Since the mass and average radius of the Moon are 7.350 x 1022 kg and 1740 km, respectively, how large must the gravitational force (in N) be that acts on Captain Cutshaw?
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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Please answer the following questions and explain how you solved the problem. Be sure to state what the known and unknown quantities are, what concepts were applied, and what equations were used. Thank you!
![**Gravitational Force Calculation on an Astronaut**
Captain Cutshaw is a 75.9 kg astronaut on board an experimental space station that orbits the Moon at an altitude of 269 km above the Moon's surface.
Since the mass and average radius of the Moon are \( 7.350 \times 10^{22} \) kg and 1740 km, respectively, how large must the gravitational force (in N) be that acts on Captain Cutshaw?
---
To find the gravitational force acting on Captain Cutshaw, we use the gravitational force formula:
\[ F = \frac{{G \cdot m_1 \cdot m_2}}{{r^2}} \]
where:
- \( F \) is the gravitational force,
- \( G \) is the gravitational constant \((6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2})\),
- \( m_1 \) is the mass of the Moon \((7.350 \times 10^{22} \, \text{kg})\),
- \( m_2 \) is the mass of the astronaut \((75.9 \, \text{kg})\),
- \( r \) is the distance between the center of the Moon and the astronaut.
**Calculations:**
1. Convert the radius and altitude to meters:
- Radius of the Moon: 1740 km = \( 1,740,000 \, \text{m} \)
- Altitude above the Moon: 269 km = \( 269,000 \, \text{m} \)
2. Calculate the total distance \( r \):
\[ r = \text{Radius of the Moon} + \text{Altitude above the Moon} \]
\[ r = 1,740,000 \, \text{m} + 269,000 \, \text{m} \]
\[ r = 2,009,000 \, \text{m} \]
3. Insert the values into the gravitational force formula:
\[ F = \frac{{(6.67430 \times 10^{-11}) \cdot (7.350 \times 10^{22}) \cdot (75.9)}}{{(2,009,000](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6094c4bd-a020-481f-869e-f65291d4ed64%2F8ed443cc-28d3-49ff-95d4-dbed78c18d96%2F7bvph4w_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Gravitational Force Calculation on an Astronaut**
Captain Cutshaw is a 75.9 kg astronaut on board an experimental space station that orbits the Moon at an altitude of 269 km above the Moon's surface.
Since the mass and average radius of the Moon are \( 7.350 \times 10^{22} \) kg and 1740 km, respectively, how large must the gravitational force (in N) be that acts on Captain Cutshaw?
---
To find the gravitational force acting on Captain Cutshaw, we use the gravitational force formula:
\[ F = \frac{{G \cdot m_1 \cdot m_2}}{{r^2}} \]
where:
- \( F \) is the gravitational force,
- \( G \) is the gravitational constant \((6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2})\),
- \( m_1 \) is the mass of the Moon \((7.350 \times 10^{22} \, \text{kg})\),
- \( m_2 \) is the mass of the astronaut \((75.9 \, \text{kg})\),
- \( r \) is the distance between the center of the Moon and the astronaut.
**Calculations:**
1. Convert the radius and altitude to meters:
- Radius of the Moon: 1740 km = \( 1,740,000 \, \text{m} \)
- Altitude above the Moon: 269 km = \( 269,000 \, \text{m} \)
2. Calculate the total distance \( r \):
\[ r = \text{Radius of the Moon} + \text{Altitude above the Moon} \]
\[ r = 1,740,000 \, \text{m} + 269,000 \, \text{m} \]
\[ r = 2,009,000 \, \text{m} \]
3. Insert the values into the gravitational force formula:
\[ F = \frac{{(6.67430 \times 10^{-11}) \cdot (7.350 \times 10^{22}) \cdot (75.9)}}{{(2,009,000
![### Physics of High-Speed U-Turns: Tension in the Cable
In a high-speed car chase involving the Diplomatic Security Service, a saboteur releases an oil slick from their vehicle and then makes a sudden U-turn. DSS agent Luke Hobbes, having only a fraction of a second to react, uses quick thinking and fires a grapple hook from the side of his car toward a nearby utility pole. Once the hook is anchored, the cable attached to it enables Hobbes to make an 8.52-meter diameter U-turn in just 3.12 seconds.
#### Problem Statement:
If the tension in the cable is the only horizontal force acting on the car, what must be the tension (in Newtons) if Hobbes and his car have a combined mass of 1,993 kilograms?
---
**Objective:**
To determine the tension in the cable during the high-speed U-turn.
**Given Data:**
- Diameter of the U-turn: 8.52 meters
- Time to complete the U-turn: 3.12 seconds
- Combined mass of Hobbes and the car: 1,993 kg
**Key Concepts:**
1. **Centripetal Force:** The force that keeps the car moving in a circular path.
2. **Tension in the Cable:** Acts as the centripetal force.
**Formulas:**
1. **Radius (r) of the U-turn:**
\[
r = \frac{Diameter}{2}
\]
2. **Centripetal Force (F):**
\[
F = \frac{mv^2}{r}
\]
3. **Velocity (v):**
\[
v = \frac{2\pi r}{T}
\]
**Step-by-step Solution:**
1. Calculate the radius of the U-turn.
\[
r = \frac{8.52\, \text{m}}{2} = 4.26\, \text{m}
\]
2. Calculate the velocity of the car.
\[
v = \frac{2\pi r}{T} = \frac{2\pi \times 4.26\, \text{m}}{3.12\, \text{s}} = 8.59\, \text{m/s}
\]
3. Calculate the centripetal force (which is provided by the tension](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6094c4bd-a020-481f-869e-f65291d4ed64%2F8ed443cc-28d3-49ff-95d4-dbed78c18d96%2Fudwtxyj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Physics of High-Speed U-Turns: Tension in the Cable
In a high-speed car chase involving the Diplomatic Security Service, a saboteur releases an oil slick from their vehicle and then makes a sudden U-turn. DSS agent Luke Hobbes, having only a fraction of a second to react, uses quick thinking and fires a grapple hook from the side of his car toward a nearby utility pole. Once the hook is anchored, the cable attached to it enables Hobbes to make an 8.52-meter diameter U-turn in just 3.12 seconds.
#### Problem Statement:
If the tension in the cable is the only horizontal force acting on the car, what must be the tension (in Newtons) if Hobbes and his car have a combined mass of 1,993 kilograms?
---
**Objective:**
To determine the tension in the cable during the high-speed U-turn.
**Given Data:**
- Diameter of the U-turn: 8.52 meters
- Time to complete the U-turn: 3.12 seconds
- Combined mass of Hobbes and the car: 1,993 kg
**Key Concepts:**
1. **Centripetal Force:** The force that keeps the car moving in a circular path.
2. **Tension in the Cable:** Acts as the centripetal force.
**Formulas:**
1. **Radius (r) of the U-turn:**
\[
r = \frac{Diameter}{2}
\]
2. **Centripetal Force (F):**
\[
F = \frac{mv^2}{r}
\]
3. **Velocity (v):**
\[
v = \frac{2\pi r}{T}
\]
**Step-by-step Solution:**
1. Calculate the radius of the U-turn.
\[
r = \frac{8.52\, \text{m}}{2} = 4.26\, \text{m}
\]
2. Calculate the velocity of the car.
\[
v = \frac{2\pi r}{T} = \frac{2\pi \times 4.26\, \text{m}}{3.12\, \text{s}} = 8.59\, \text{m/s}
\]
3. Calculate the centripetal force (which is provided by the tension
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