Environmental Engineering Reference
In-Depth Information
9.5 First part is obtained by equating the gravitational attraction to the centripetal
force. After the dust cloud has been driven away, the total energy of the
planet is
4
3
πρr
3
)m
r
G(M
+
1
2
GMm
r
−
and the orbit is bound if this is negative, hence the result. Determine
semi-major axis by equating the total energy to
−
GMm/(
2
a)
. The answer
M
dust
/M)
−
1
4
3
πρr
3
.
At distance
r
the velocity is tangential and so the planet is at pericentre
(
r<a
so cannot be apocentre). Thus
a(
1
can be re-arranged to read
a
=
r(
1
−
where
M
dust
=
−
ε)
=
r
and so
ε
=
M
dust
/M
.
9.6 Use
u
2
1
)GM/a
to obtain the speed immediately after the firing
of the rockets (apocentre of the Eagle's orbit). We have that
r
=
(
2
a/r
−
=
1850 km
but need
a
. Solve for
a
using
a(
1
−
ε)
=
1755 km and
a(
1
+
ε)
=
1850
km. Hence speed is 1.61 km s
−
1
which implies a slowing down of around
20 m s
−
1
.
9.7 The difference is due to the fact that the Sun is offset from the centre of
the Earth's elliptical orbit by a distance
εR
. The summertime orbit therefore
sweeps out an area
4
εR
2
more than the wintertime orbit. Since the area
swept out over the whole year is
≈
πR
2
and since area is swept out at a
constant rate this translates into a time difference of 4
ε/π
years
≈
≈
8days.
This compares well with the 7 days difference between the period 21 March
to 22 September and the period 23 September to 20 March.
PROBLEMS 10
10.1 The hoop receives an impulse
p
on the rim which implies an angular
impulse
L
=
r
×
p
about the centre of the hoop. The angular momentum
before impact is
mr
2
ω
n
,
L
=
I
ω
=
where
n
is horizontal. To only deflect the hoop
L
must also lie in the
horizontal plane, i.e.
r
must point up and the blow should be applied to the
highest point of the hoop. A blow anywhere else will induce a wobble. The
change in direction
θ
is
rp
mr
2
ω
=
p
mv
,
θ
≈
tan
θ
=
rω
.
10.2 Compute the torque about the centre of mass because the coin accelerates
(otherwise we would have to include the effect of fictitious forces). The
torque is a result of the normal force and friction at the base of the coin.
The normal force is equal in magnitude to the weight and friction provides
wherewehaveused
v
=
