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5 Gravitation
−
Gm M
√
2R
=
1
2
mv
2
−
Gm M
R
∴ v =
(2 −
√
2)
G M
R
5.25 The potential energy of the satellite on the earth’s surface is
U (R) = −
G Mm
R
(1)
where M and m are the mass of the earth and the satellite, respectively, and R
is the earth’s radius.
The potential energy at a height h = 0.5R above the earth’s surface will be
U (R + h) = −
G Mm
R + h
= −
G Mm
1.5R
(2)
Gain in potential energy
U = −
G Mm
1.5R
−
−
G Mm
R
=
G Mm
3R
(3)
Thus the work done W 1 in taking the satellite from the earth’s surface to a
height h = 0.5 r
W 1 =
G Mm
3R
(4)
Extra work W 2 required to put the satellite in the orbit at an attitude h = 0.5R
is equal to the extra energy that must be supplied:
W 2 =
1
2
mv 0 =
1
2
m
G M
R + h
=
G Mm
3R
(5)
where v 0 is the satellite’s orbital velocity.
Thus from (4) and (5),
W 1
W 2
= 1.0.
5.26 The initial angular momentum of the asteroid about the centre of the planet is
L = mv 0 d.
At the turning point the velocity v of the asteroid will be perpendicular to
the radial vector. Therefore the angular momentum L = mv R if the asteroid
is to just graze the planet. Conservation of angular momentum requires that
L = L . Therefore