A projectile is shot straight up from the earth’s surface at a speed of 1.50×10^4 km/hr.?

1 Answer

• first convert this speed to m/s; 1.5×10^4km/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=6.67×10^(-11)in MKS units

  M=6×10^24kg

  we get (1/Re-1/r) = 2.17×10^(-8)

  Re=6.4×10^6 m so 1/r = 1.34×10^(-7) so r=7.43×10^6m

  this is the distance from the center of the earth, or (7.43-6.4)x10^6=1.03×10^6 m

  above the surface of the earth; this is 1031 km above the surface of the earth