**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 GravityGravity 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 GravityEarth's surface gravity \( g_E \) is approximately 9.81 \( m/s^2 \).### Titan’s Surface GravityTitan 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 km2. **Mass and Radius of Earth**: - Mass of Earth (\( M_E \)): \( 5.97237 \times 10^{24} \) kg - Radius of Earth (\( r_E \)): 6,371 km### Calculation1. 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

Gravity is an attractive force. It exists between all objects that have a mass. The force of…

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Show that Titan's surface gravity is about one-seventh that of Earth's.