Show that Titan's surface gravity is about one-seventh that of Earth's.
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![**Demonstrating Titan's Surface Gravity in Relation to Earth's**
To show that Titan’s surface gravity is about one-seventh that of Earth’s, we need to compare the gravitational forces on the surface of both celestial bodies.
### Understanding Gravity
Gravity is the force that attracts a body toward the center of the Earth, or any other physical body having mass. The surface gravity of a celestial body is dependent on its mass and radius. The formula to calculate surface gravity (\( g \)) is given by:
\[ g = \frac{GM}{r^2} \]
Where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the celestial body,
- \( r \) is the radius of the celestial body.
### Earth's Surface Gravity
Earth's surface gravity \( g_E \) is approximately 9.81 \( m/s^2 \).
### Titan’s Surface Gravity
Titan is the largest moon of Saturn, and its physical characteristics contribute to its lower surface gravity. Titan has a mass \( M_T \) and a radius \( r_T \) which are different from those of Earth.
1. **Mass and Radius of Titan**:
- Mass of Titan (\( M_T \)): \( 1.3452 \times 10^{23} \) kg
- Radius of Titan (\( r_T \)): 2,575 km
2. **Mass and Radius of Earth**:
- Mass of Earth (\( M_E \)): \( 5.97237 \times 10^{24} \) kg
- Radius of Earth (\( r_E \)): 6,371 km
### Calculation
1. Using the formula for surface gravity:
\[ g_T = \frac{GM_T}{r_T^2} \]
2. Substitute the respective values for Titan:
- \( G = 6.67430 \times 10^{-11} \) m\(^3\)kg\(^{-1}\)s\(^{-2}\)
- \( M_T = 1.3452 \times 10^{23} \) kg
- \( r_T = 2,575,000 \) m
\[ g_T = \frac{(6.67430 \times 10^{-11})(1.3452 \times 10^{23})}{(2,575,000)^2} \]
Performing the calculation](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffc44715e-e997-4081-aac6-58f75fd336c7%2F68f72c0f-19dc-489c-a745-e4e16068ab86%2Fgqk0znn_processed.png&w=3840&q=75)
Transcribed Image Text:**Demonstrating Titan's Surface Gravity in Relation to Earth's**
To show that Titan’s surface gravity is about one-seventh that of Earth’s, we need to compare the gravitational forces on the surface of both celestial bodies.
### Understanding Gravity
Gravity is the force that attracts a body toward the center of the Earth, or any other physical body having mass. The surface gravity of a celestial body is dependent on its mass and radius. The formula to calculate surface gravity (\( g \)) is given by:
\[ g = \frac{GM}{r^2} \]
Where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the celestial body,
- \( r \) is the radius of the celestial body.
### Earth's Surface Gravity
Earth's surface gravity \( g_E \) is approximately 9.81 \( m/s^2 \).
### Titan’s Surface Gravity
Titan is the largest moon of Saturn, and its physical characteristics contribute to its lower surface gravity. Titan has a mass \( M_T \) and a radius \( r_T \) which are different from those of Earth.
1. **Mass and Radius of Titan**:
- Mass of Titan (\( M_T \)): \( 1.3452 \times 10^{23} \) kg
- Radius of Titan (\( r_T \)): 2,575 km
2. **Mass and Radius of Earth**:
- Mass of Earth (\( M_E \)): \( 5.97237 \times 10^{24} \) kg
- Radius of Earth (\( r_E \)): 6,371 km
### Calculation
1. Using the formula for surface gravity:
\[ g_T = \frac{GM_T}{r_T^2} \]
2. Substitute the respective values for Titan:
- \( G = 6.67430 \times 10^{-11} \) m\(^3\)kg\(^{-1}\)s\(^{-2}\)
- \( M_T = 1.3452 \times 10^{23} \) kg
- \( r_T = 2,575,000 \) m
\[ g_T = \frac{(6.67430 \times 10^{-11})(1.3452 \times 10^{23})}{(2,575,000)^2} \]
Performing the calculation
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