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Calculate the gravitational potential due to a thin rod of length 2L and mass M at a distance R from the center of the rod and in a direction perpendicular to the rod Select one: 1100 -GM In(+√4+R²) L 0 410 GM AL 16=0 +√(+R²) In(+ √(²²+R²) =+√4/+R²) L+ √(L² + R²) -L+ √(L² + R²) =GM In(+√(4+R²) +√²+R²) L GM In( 2L

Calculate the gravitational potential due to a thin rod of length 2L and mass M at a distance R from the center of the rod and in a direction perpendicular to the rod Select one: 1100 -GM In(+√4+R²) L 0 410 GM AL 16=0 +√(+R²) In(+ √(²²+R²) =+√4/+R²) L+ √(L² + R²) -L+ √(L² + R²) =GM In(+√(4+R²) +√²+R²) L GM In( 2L

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Calculate the gravitational potential due to a thin rod of length 2L and mass M at a distance R from the center of the rod and in a direction perpendicular to the rod Select one: GM 6=- In(- + √(4/+R²) +√²+R²) +√4/+R²) =+√(¹+R²) 06=0 -In(- GM AL GM 0 =- In( 2L GM $=-- In( L L+ √(L² + R²) -L+ √(L² + R²) +√4/+R²) ==++√(²²+R²)
Transcribed Image Text:Calculate the gravitational potential due to a thin rod of length 2L and mass M at a distance R from the center of the rod and in a direction perpendicular to the rod Select one: GM 6=- In(- + √(4/+R²) +√²+R²) +√4/+R²) =+√(¹+R²) 06=0 -In(- GM AL GM 0 =- In( 2L GM $=-- In( L L+ √(L² + R²) -L+ √(L² + R²) +√4/+R²) ==++√(²²+R²)
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