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It is known that the mass of the sun is 1.97 * 10 ^ 30kg, the mass of the earth is 6.68 * 10 ^ 24kg, and the average distance between the sun and the earth is 1.49 * 10 ^ 11m The gravitational force between the sun and the earth is_____ N. It is known that the tensile force of 5.98 * 10 ^ 24N is needed for a steel bar with a cross-sectional area of 1cm, so what is the cross-sectional area that can be pulled by the gravitational force between the earth and the sun_____ M ^ 2 Please write down the analysis process,
According to the law of gravitation, f = GMM / R ^ 2, g = 6.67 × 10 ^ - 11 n · m2 / kg2. Then we can calculate f = 6.67 × 10 ^ - 11 * 1.97 * 10 ^ 30 * 6.68 * 10 ^ 24 / [1.49 * 10 ^ 11] ^ 2 = 3.95 * 10 ^ 22 n
The mass of the earth is 6x10 ^ 24kg, and the distance between the earth and the sun is 1.5x10 ^ 11. It is regarded as a uniform circular motion. How much is the sun's gravity on the earth
Because f = GMM / R ^ 2 = M4 π ^ 2R / T ^ 2
Because t = 365 days
So f = 6x10 ^ 24x4 π ^ 2 / (365x86400) ^ 2 = self calculated
The mass of the sun is 2 * 10 ^ 30kg, and the distance between the earth and the sun is 1.5 * 10 ^ 11?
You didn't provide the mass of the earth, so, you go back to check the mass of the earth, and then use the law of gravitation f = gm1m2 / R2 to calculate, the result is 3.56 * 10 to the 22nd power n (data from the current physics textbook of the people's education, Vol. 1, Chapter 6, page 107), which is more accurate
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