In: Physics

A 150-kg satellite is in circular orbit of radius 7.3 Mm around Earth. Determine:(a) potential, kinetic, and total mechanical energies.(b) the orbital speed.(c) the escape velocity from this altitude.

a)

Potential energy, PE = - GMm/R

Where G is the universal gravitational constant, M is the mass of
earth, m is the mass of satellite and R is the orbital radius of
the satellite.

PE = - [6.674 x 10^{-11} x 5.972 x 10^{24} x 150] /
(7.3 x 10^{6})

= - 8.19 x 10^{9} J

b)

Force between the satellite and earth, F = GMm/R^{2} =
mv^{2}/R

Where mv^{2}/R is the centripetal force

GMm/R =
mv^{2}
[Multiplying with R on both the sides]

Kinetic energy = 1/2 mv^{2} = 1/2
GMm/R
[Multiplying with 1/2 on both the sides]

= 1/2 x PE

= 1/2 x 8.19 x 10^{9} J

= 4.095 x 10^{9} J

c)

Total mechanical energy = KE + PE

= 1/2 GMm/R + [-GMm/R]

= - 1/2 GMm/R

= - 4.095 x 10^{9} J

d)

From part b, 1/2 mv^{2} = 4.095 x 10^{9} J

v^{2} = ( 2 x 4.095 x 10^{9}) /150

v = SQRT[( 2 x 4.095 x 10^{9}) /150]

= 7.39 x 10^{3} m/s

e)

When the satellite attains escape velocity, its total energy will
be zero.

KE + PE = 0

1/2 mv^{2} - GMm/R = 0

1/2 mv^{2} = GMm/R

= 8.19 x
10^{9}
[From part a]

v^{2} = ( 2 x 8.19 x 10^{9}) /150

v = SQRT[( 2 x 8.19 x 10^{9}) /150]

= 10.45 x 10^{3} m/s

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