The force two bodies with mass exert on each other is given by the gravitational

Question

The force two bodies with mass exert on each other is given by the gravitational force: F = Gm1m1/r2. Where G is a constant (look up!) and r is the separation between the centers of the two bodies. The gravitational potential energy of a two-body system is likewise given by: U = Gm1m2/r.

a) Show that on the surface of the Earth, this expression really does simplify to: F = mg, where g = 9.8 m/s2.

b) The find an expression for the minimum velocity needed to escape the gravitational pull of the Earth. To do this, use conservation of energy and consider two snapshots: launching on the surface of the Earth, and infinitely far from the Earth.

Explanation / Answer

a) weknow, F = -dU/dr

= -( -G*m1*m2*(-1*r^(-1-1))

= -G*m1*m2/r^2

let m1 = M_earth, m2 = m

F = -G*M_earth*m/r^2

= -m*(G*M_earth/r^2)

= -m*g

b) Apply conservatio of energy

initial mechanical energy = final mechinical energy

U1 + K1 = U2 + K2

-G*Me*m/r + (1/2)*m*ve^2 = 0 + 0

(1/2)*m*ve^2 = G*Me*m/r

ve^2= 2*G*Me/r

ve = sqrt(2*G*Me/r)

w know, Me = 5.98*10^24 kg

r = 6370 km = 6.37*10^6 m

so,

eccape speed on the surface of the earth, ve = sqrt(2*6.67*10^-11*5.98*10^24/(6.37*10^6))

= 11190 m/s

= 11.19 km/s

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