At the same instant, a 85.2 kg satellite is in circular orbit around the Earth a
ID: 1482365 • Letter: A
Question
At the same instant, a 85.2 kg satellite is in circular orbit around the Earth at a height of 2.50RE, a 61416 kg Boeing 737 jet crosses the Atlantic at 2.20 × 102 m/s and a nimble 0.664 g fly moves at 12.4 m/s. This is all occurring while you are driving your 3091 lb vehicle 63.5 mph home from work. Using the ranking module below, sort the four objects by decreasing kinetic energy, also enter the kinetic energy of the satellite, KEsat, in the answer module below. The gravitational constant is G = 6.67 × 10-11 N·m2/kg2, the mass of the Earth is mE = 5.98 × 1024 kg, and the radius of the Earth is RE= 6.37 × 106 m.
t, a 85.2 kg sat kg Boeing 737 jet crosses the Atlantic at 2.20 x 102 m/s and a nimble 0.664 g fly moves at 12.4 m/s. This is all occurring while you are driving your 3091 lb vehicle 63.5 mph home from work. Using the ranking module below, sort the four objects by decreasing kinetic energy, also enter the kinetic energy of the satellite, KEsat, in the answer module below. The gravitational constant is G 6.67 x101 N m2/kg2, the mass of the Earth is mE 5.98 x1024 kg, and the radius of the Earth is RE 6.37 x105 m. Most Kinetic Energy Least Kinetic Energy car jet fly satellite Number KEsatExplanation / Answer
For "orbit", centripetal acceleration = gravitational acceleration
v2/r=GM/r2 which rearranges to v2=GM/r
where G = Newton's gravitational constant = 6.674*1011 N·m²/kg²
and mass of the Earth is mE = 5.98*1024 kg
and r = orbit radius = 2.50RE
r= 2.5*6.371*106 m = 15.9275*106 mRE=1.59*106 m
v² = GM / r
=G*mE/r
= 6.674*1011 N·m²/kg² * 5.98*1024 kg/15.9275*106 mRE
v²=2.50*107Nm/KgRE or 25*106m²/s4
KE = ½mv²= ½ * 85.2kg * 25*106m²/s4
=1065*106kgm²/s4=1.65 GJ
The airplane has KE = ½mv²
= ½ * 61416kg* (2.20 × 102 m/s )²
= 148626.72*106 kgm²/s4=1.48GJ
so the ranking, in decreasing kinetic energy
jet, satellite, car, fly
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