Answer:
a. v = 1.679 × 10⁹ m/s b. The answer of v = 1.679 × 10⁹ m/s is not reasonable
Explanation:
Here is the complete question
You read in a science magazine that on the Moon, the speed of a shell leaving the barrel of a modern tank is enough to put the shell in a circular orbit above the surface of the Moon (there is no atmosphere to slow the shell).
A.....What should be the speed for this to happen? Assume that the radius of the Moon is rM = 1.74×106m, and the mass of the Moon is mM = 7.35×1022kg.(Express your answer to two significant figures and include the appropriate units.)
B.....Is this number reasonable?
Solution
From the question, the centripetal force in orbit for the shell = gravitational force of moon on shell.
So, mv²/r = GMm/r²
So v = √(GM/r) where v = velocity of shell, G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², M = mass of moon = 7.35 × 10²² kg and r = radius of moon = 1.74 × 10⁶ m.
v = √(GM/r) = v = √(6.67 × 10⁻¹¹ Nm²/kg² × 7.35 × 10²² kg/1.74 × 10⁶ m)
v = √(49.0245/1.74 × 10¹⁷) m/s
v = √28.175 × 10¹⁷ m/s
v = √2.8175 × 10¹⁸ m/s
v = 1.679 × 10⁹ m/s
b. The answer of v = 1.679 × 10⁹ m/s is not reasonable because, it is over 1000000 km/s which is greater than the speed of light which is 300000 km/s
