QQuestionAnatomy and Physiology
QuestionAnatomy and Physiology
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Step 1I.
g_{resultant} = g_{Earth} + g_{moon} = 2.68 \times 10^{-3} \text{ m/s}^2 + 1.58 \times 10^{-5} \text{ m/s}^2 = 2.69 \times 10^{-3} \text{ m/s}^2
Understanding the Problem The problem asks us to calculate the period of rotation of the moon about the Earth, given the radius of the moon's orbit as 3.5 x 10^6 km and the mass of the Earth as 60 x 10^24 kg. II. Calculating the Period of Rotation To calculate the period of rotation, we can use Kepler's Third Law, which states that the square of the period of a planet is proportional to the cube of the semi-major axis of its orbit. The formula for Kepler's Third Law is: First, we need to convert the radius of the moon's orbit from kilometers to meters: III. Calculating the Period of Rotation (continued) Therefore, the period of rotation of the moon about the Earth is approximately 5.43 million seconds. IV. Calculating the Resultant Gravitation Field The problem also asks us to calculate the resultant gravitation field of the Earth and the moon at a point on the line joining the centers of the Earth and the moon. The formula for the gravitational field of a spherical body is: The gravitational field of the Earth is: The gravitational field of the moon is: V. Calculating the Resultant Gravitation Field (continued) Assuming the mass of the moon is 7.35 x 10^22 kg and the distance from the center of the moon is 3.85 x 10^8 m, we can calculate the gravitational fields of the Earth and the moon: Therefore, the resultant gravitational field is: VI.
Final Answer
The period of rotation of the moon about the Earth is approximately 5.43 million seconds. The resultant gravitational field of the Earth and the moon at a point on the line joining the centers of the Earth and the moon is approximately 2.69 x 10^- 3 m/s^2.
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