mean° 5.9742 x 1024 kg earth -6.3781x 10%m mngon 7.36 x 10*4 kg moon -1.7374x 10

Question

mean° 5.9742 x 1024 kg earth -6.3781x 10%m mngon 7.36 x 10*4 kg moon -1.7374x 10%m earth to mon G 6.67428 x 101 N-m2/kg A 1900 kg satellite is orbitting the earth in a circular orbit with an altitude of 1400 km. 3.844 x 10 m (center to center) 1) How much energy does it take just to get it to this altitude? Submit 2) How much kinetic energy does it have once it has reached this altitude? Submit 3) What is the ratio of the this change in potential energy to the change in kinetic energy? (i.e. what is (a)/(b)2) Submit 4) What would this ratio be if the final altitude of the satellite were 4100 km? Submit 5) What would this ratio be if the final altitude of the satellite were 3185 km? Submit

Explanation / Answer

given Me = 5.9742*10^24

Re = 6.3781*10^6 m

Mm = 7.36*10^22 kg

Rm = 1.7374*10^6 m

d = 3.844*10^8 m

G = 6.67428*10^-11 Nm^2/kg^2

m = 1900 kg

a. energy = 0.5mv^2 + GmMe/(Re + h)

also

0.5mv^2/(Re + h) = GmMe/(Re + h)^2

hence

E = 2GmM/(Re + h) = 2*6.67428*5.9742*1900*10^(24-11)/(6.3781*10^6 + 1400,000) = 194802377944.22802 J = 194802.37794 MJ

b. KE = 0.5mv^2 = GmMe/(Re + h) = E/2 = 97401.18897 MJ = 97401.18897*10^6 J

c. a/b = 2

d. for any altitude this ratio is the same

e. for any altitude this ratio is the same hence the altitude does not matter

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