Mercury has a mass MP = 3.18 x 1023 kg, an orbital period T = 88.0 days, and an
ID: 2145803 • Letter: M
Question
Mercury has a mass MP = 3.18 x 1023 kg, an orbital period T = 88.0 days, and an orbital radius r = 5.79 x 1010 m. Assume the orbit of Mercury around the Sun is approximately circular.a)Find the angular speed v(t) of Mercury about the Sun.
I'VE DONE THIS ALREADY
Angular speed (w)=2pi/T=2pi/88=0.0714 rad/day = 8.3E-7 rad/sec
b)What is the linear speed vt of Mercury?
I'VE DONE THIS ALREADY
linear speet v(t)=rw=(5.79E10)(8.26E-7)=4.78E4 m/s
c)Determine the centripetal acceleration aC of Mercury due to its orbit about the Sun?
I'VE DONE THIS ALREADY
Ac = V^2/r...(4.78E4)^2/(5.79E10)=3.95E-2 m/s^2
d)Calculate the centripetal force FC acting on Mercury which causes its circular orbit about the Sun? Use Newton
Explanation / Answer
PE = -G(Msun)(Mmercury)/R
R = 5.79 x 1010 m
PE = - 6.673 x 10-11 x 1.992 x 1030 x 3.18 x 1023 / 5.79 x 1010 = -7.3 x 1032 Joules
Total Energy = KE + PE = -3.53 x 1032 Joules