What is an oxygen molecule\'s escape velocity from the \'\'top\'\' of the Earth\

Question

What is an oxygen molecule's escape velocity from the ''top'' of the Earth's atmosphere (called the Karman line) at 100. km high (62.1 mi) if it's total mass is 32.0 amu = 5.31x10^-26 kg (ME = 5.97x10^24 kg, RE = 6370 km)? (b) What is the gravitational potential energy of an O2 molecule at the Karman line? Apply conservation principles to find the initial speed vi of a 10.5 g bullet shot into a 2.55 kg block attached to a spring with a force constant of 245 N/m, given that the max compression of the spring caused by the bullet is 17.5 cm. If we can approximate a grandfather clock as a simple pendulum then what length will its pendulum have to be to keep accurate time (note, it should take exactly 1.00 s to swing half a period, show your work ...)?

Explanation / Answer

here

m*v2 / 2 = (G*M*m) / R

Where m is the mass of the object, M mass of the earth, G is the gravitational constant, R is the radius of the earth, and v is the escape velocity. It simplifies to:

v = sqrt(2*G*M / R)

v = sqrt{(2 * 6.673 * 10^-11 * 5.97 * 10^24) / 6370)}

v = 12.50 * 10^10 m/s

the escape velocity is 12.50 * 10^10 m/s

b)

The Gravitational potential energy is U = G*M*m / r
                                                       = (6.673 * 10^-11 * 5.31 * 10^-26 * 5.97 * 10^24) / 100
                                                       = 2.12 * 10^-13 J

GPE = 2.12 * 10^-13 J

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