please do this problem from gravitation
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behemoth with a radius of 65,000 kilometers. When the starship is 2,500 kilometers from the planet’s surface, what is the starship’s acceleration (providing that its engines are turned off)? Round your answer to two significant digits.
Determine the centripetal acceleration of the moon. (Assuming the moon is held in it's orbit by the gravitational force of the Earth, you are then also calculating the acceleration due to gravity of the Earth at the moon's orbit.)
Determine the ratio of the radius of the moon's orbit to the radius of the Earth.
Use the results of a. and b. to calculate the acceleration due to gravity on the surface of the Earth.
How does this value compare to the generally accepted value of g? Are the results of your calculations in close enough agreement with experimental observations to verify the inverse square rule for gravitation? Discuss briefly.
Estimate the value of the universal gravitational constant from the following approximate measurements taken during the original Cavendish experiment (and converted into SI units for us)…
one hundred kilogram fixed and one kilogram rotating masses
ten centimeter separation between fixed and rotating masses
one millionth newton of force on each of the rotating masses
Determine the centripetal acceleration of the moon. (Assuming the moon is held in it's orbit by the gravitational force of the Earth, you are then also calculating the acceleration due to gravity of the Earth at the moon's orbit.)
Determine the ratio of the radius of the moon's orbit to the radius of the Earth.
Use the results of a. and b. to calculate the acceleration due to gravity on the surface of the Earth.
How does this value compare to the generally accepted value of g? Are the results of your calculations in close enough agreement with experimental observations to verify the inverse square rule for gravitation? Discuss briefly.
Estimate the value of the universal gravitational constant from the following approximate measurements taken during the original Cavendish experiment (and converted into SI units for us)…
one hundred kilogram fixed and one kilogram rotating masses
ten centimeter separation between fixed and rotating masses
one millionth newton of force on each of the rotating masses
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