16. (a) Near the surface of the Earth, the gravitational force is constant: Fg =

Question

16. (a) Near the surface of the Earth, the gravitational force is constant: Fg = mg. Using conservation of energy, find an algebraic expression for the maximum height, h, that an object launched upwards at velocity v can reach before falling back towards Earth's surface.

(b) In general, however, the gravitational force between two objects of mass M and m depends on the distance between these objects, r. The magnitude of this force is given by Fg = GMm=r2, where G is the universal gravitational constant. In this case, the gravitational potential energy is given by Ug = -GMm/r:

For an object of mass m on the surface of a planet (M), the minimum speed required to escape the planet's gravitational field is known as the escape velocity, vE. For an object initially a distance r from the center of the planet, find vE. (Your expression should involve some or all of the variables M, m, r, and G.)

(c) Using your result above, calculate the escape velocity for an object on Earth's surface.

Explanation / Answer

a)
use conservation of energy
Kinetic energy at bottom = potential energy at maximum height
0.5*m*v^2 = m*g*h
0.5*v^2 = g*h
h = v^2/(2g)

b)
initial sum of kinetic energy + potential energy should be 0 as final energy is 0 when object has escaped
0.5*m*v^2 - G*M*m/r = 0
v = sqrt (2*G*M/r)

c)
for earth:
M= 5.972*10^24 kg
r = 6.371*10^6 m
v = sqrt (2*G*M/r)
= sqrt (2* 6.673*10^-11 * 5.972*10^24/(6.371*10^6))
= 11185 m/s

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