Question

A projectile is shot straight up from the Earth's surface at a speed of 1.5 * 10^4 km/hr. How high does it go?

Answer 1

How did you get that answer?

Answer 2

first convert this speed to m/s; 1.5x10^4km/hr= 4166.7m/s

the speeds involved in this problem will cause the projectile to rise high enough that we cannot regard gravity as a constant, so we can't use g=9.8m/s/s since the projectile will rise high enough such that the value of g will vary significantly during the projectile's trajectory

we can use the formulation of conservation of energy in which we define PE as -GMm/r where G is the newtonian grav cst, M is the mass of the earth, m is the mass of the projectile and r is the distance of the projectile from the center of the earth

on the surface of the earth, the projectile has KE of 1/2 mv^2 and PE of -GMm/Re where Re is the radius of the earth

at its apex, v=0 and PE =-GMm/r where r is the distance of the highest point from the center of the earth.

equating energies, we have

1/2 mv^2 - GMm/Re = - GMm/r or

v^2 = 2GM(1/Re-1/r)

using v=4.17 x10^3m/s
G=6.67x10^(-11)in MKS units
M=6x10^24kg

we get (1/Re-1/r) = 2.17x10^(-8)

Re=6.4x10^6 m so 1/r = 1.34x10^(-7) so r=7.43x10^6m

this is the distance from the center of the earth, or (7.43-6.4)x10^6=1.03x10^6 m
above the surface of the earth; this is 1031 km above the surface of the earth

Answer 3

its 4166.7m/s

Answer 4

Thank you soo much! That was a perfect explanation =]