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Physic-Moment of inertia
We assumed that the Mars is approximately three times as far away from the Sun as the Mercury and has one-sixth of the Mercury's mass. What is the ratio of the monent of inertia of the Mars to that of the Mercury, about an axis through the Sun, perpendicular to the plane of the Solar System?
最佳解答:
Your assumptions have deviated from the reality quite a lot... But anyway, I will calculate the answer according to the given data. Let Ir, Iy be the moment of inertia of Mars to that of Mercury about an axis through the Sun, perpendicular to the plane of the Solar System. Ratio of mass of Mars to that of Mercury, Mr : My = 1 : 6 Ratio of distance of Mars to that of Mercury from the Sun, rr : ry = 3 : 1 Assuming Mars and Mercury are spherically symmetric, so that we can assume the C.G. of the planets are from the geometric centre. By I = Mr2 Required answer = Ir / Iy = Mrrr2/ Myry2 = (Mr / My)(rr / ry)2 = (1 / 6)(3 / 1)2 = 3 / 2 The required ratio is 3 : 2.
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Physic-Moment of inertia
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發問:We assumed that the Mars is approximately three times as far away from the Sun as the Mercury and has one-sixth of the Mercury's mass. What is the ratio of the monent of inertia of the Mars to that of the Mercury, about an axis through the Sun, perpendicular to the plane of the Solar System?
最佳解答:
Your assumptions have deviated from the reality quite a lot... But anyway, I will calculate the answer according to the given data. Let Ir, Iy be the moment of inertia of Mars to that of Mercury about an axis through the Sun, perpendicular to the plane of the Solar System. Ratio of mass of Mars to that of Mercury, Mr : My = 1 : 6 Ratio of distance of Mars to that of Mercury from the Sun, rr : ry = 3 : 1 Assuming Mars and Mercury are spherically symmetric, so that we can assume the C.G. of the planets are from the geometric centre. By I = Mr2 Required answer = Ir / Iy = Mrrr2/ Myry2 = (Mr / My)(rr / ry)2 = (1 / 6)(3 / 1)2 = 3 / 2 The required ratio is 3 : 2.
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