An instrument package is projected vertically upward to collect data near the to

Question

An instrument package is projected vertically upward to collect data near the top of the Earth's atmosphere (at an altitude of about 990 ). -Part A- What initial speed is required at the Earth's surface for the package to reach this height? -Part B- What percentage of the escape speed is this initial speed?

Explanation / Answer

Assuming no friction, total mechanical energy (kinetic energy plus potential energy) is conserved. The potential energy at a distance r>R from the center of the earth (R is the earth's radius) is: U = - G M m / r , where M is the earth's mass, m the package's mass, and G is the gravitational constant............... At r=R (surface), the velocity is v............................. At r = R+h, the velocity is zero (highest point reached).................................. So ,,,,, 1/2 m v^2 - G M m/R = -G M m / (R+h)................ v^2 = 2 G M (1/R - 1/(R+h) ).................. Since G M /R^2 = g ...................... v^2 = 2 g R ( 1 - R / (R+h) )........................... just substitute the values g = 9.81 m/s^2 , R = 6.36*10^6 m and h = 9.9*10^5 m to find v^2 and hence v. You will find.................. v = 4.09 km / s...................... Note that if you let h->infinity, you will have found the escape velocity (Vesc) as Vesc = sqrt(2 g R).................... So v/Vesc = sqrt(1 - R/(R+h)) , which you can express as a percentage................... This way you obtain v/Vesc = 36.68 %

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