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?

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

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.