Mercury has a mass of 3.30 1023 kg and a radius of 2.44 106 m. Assume it is a un

Question

Mercury has a mass of 3.30 1023 kg and a radius of 2.44 106 m. Assume it is a uniform solid sphere. The distance of Mercury from the Sun is 5.79 1010 m. (Assume Mercury completes a single rotation in 1.41 103 hours and orbits the Sun once every 88.0 Earth days.) What is the rotational kinetic energy of Mercury on its axis? What is the rotational kinetic energy of Mercury in its orbit around the Sun? Mercury is now revolving around the Sun at the center of the orbit. The axis of rotation is now through the center of the orbital path and perpendicular to the plane of the orbit. J

Explanation / Answer

I1 = (2/5)*M*R^2 = (2/5)*3.3*10^23*((2.44*10^6)^2)= 785.875e+33 kg m^2

W = (2*pi)/T = (2*3.14)/(1.41*1000*60*60)= 1.237e?6 rad/s

KE1 = 0.5*I1*W^2 = 601.45e+21 J

I2 = (2/5)M*R^2 + M*r^2 = 1.106e+45 kg m^2

W2 = (2*pi)/T2 = (2*3.14)/(88*24*60*60)= 825.968e?9 rad/s

KE2 = 0.5*I2*W^2 = 377.269e+30 J

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