Consider a projectile of constant mass m being fired vertically from earth (a) U
ID: 2963244 • Letter: C
Question
Consider a projectile of constant mass m being fired vertically from earth
(a) Use Newton's law of gravitation to show that the motion of the projectile, under Earth's gravitational force is governed by the equation
( rac{dv}{dx} = - rac{gR^{2}}{r^{2}})
Where R is the radius of the Earth, and g = GM/R^2
(b) Solve the differential equation from part (a) with an initial Velocity of vo
(d) If g = 9.81 m/sec^2 and R = 6370 km for Earth, what is Earth's escape velocity?
(e) If tge acceleration due to gravity for the Moon is gm = g/6 and the radius of the Moon is Rm = 1738 km, what is the escape velocity of the Moon?
Explanation / Answer
net force acting on the body is gravitational force towards center of earth
F = GMm/r^2
ma = GMm/r^2.
a = GM/r^2
a = dV/dt = -dV/dx . dx/dt = GM/r^2
V.dV/dx = -GM/r^2
we know that GM/R^2 = g
V.dV/dx = -gR^2/r^2. this is the equation
part 2 :
V.dV = -gR^2/r^2 . dX
integrating it with V from Vo to V and r changes from R to infinity
(V^2 - Vo^2)/2 = gR^2 /R = gR
V^2 = Vo^2 + 2gR
part 3: value of Vo for lim r--->infinity V^2 is applied while intergration which gives
Vo = squrt (2gR) which is the formulae for escape velocity of earth
part4:
Vo = squrt (2x9.8x6400) = 11.2 km/sec
part 5 :
Vm / Vo = squrt (g/6 x 1738) / squrt (g x 6400)
Vm = 2.383 km/sec