Download Question (15a0002) Solution

Question (15a0002)
The work needed to launch the Hubble Space Telescope into orbit.
The Hubble Space Telescope (HST or the Hubble") has a mass of about 11,100 kg and is in a nearly circular
orbit of altitude 560 km.
(a) Use the momentum principle to calculate the speed of the HST.
(b) What is the period of its orbit?
(c) What is the work needed to put the HST into orbit, starting at the surface of Earth?
Solution
(a)
Apply the
Momentum Principle.
Dene the system as the HST. Identify the forces on the HST and
sketch a free-body diagram for the HST. The only force on the HST is the gravitational force by Earth that
acts radially, toward Earth as shown below.
Figure 1: The orbit of HST.
According to the momentum principle, the net force perpendicular to the path of an object is
Fnet,rad
=
mv 2
R
where R is the radius of its path. Now, sum the forces acting in the radial direction on the object. In this
case, there is only one force, the gravitational force by Earth. Substitute this for the net force.
Fgrav
Fnet,rad
=
by Earth on HST
=
GMEarth mHST
R2
=
mv 2
R
mv 2
R
mv 2
R
Now, solve for the speed of the orbit. Note that the mass of the telescope cancels out of the equation.
GMEarth
mHST
R2
GMEarth
R
v
mv 2
R
=
= v2
r
=
GMEarth
R
Substitute the mass of the telescope and the radius of its orbit. Note that we are given its altitude, measured
from Earth's surface. Its orbital radius is measured from the center of Earth which is the radius of Earth
plus its altitude. Be sure to convert km to m.
R
= REarth + h
=
6.371 × 106 m + 5.6 × 104 m
=
6.427 × 106 m
The orbital speed is
r
v
=
GMEarth
R
s
(6.67 × 10−11 Nm2 /kg2 )(5.97 × 1024 kg)
6.427 × 106
= 7871 m/s
≈ 7900 m/s
=
(b)
The orbital period is found using the denition of average speed for an object moving in circular motion.
v
T
2πR
T
2πR
=
v
2π(6.427 × 106 m)
=
7870 m/s
= 5130 s
=
Convert the period to hours.
T
=
(5130 s)
=
1.4 h
1h
3600 s
(c)
To nd the work done on the HST to put it into orbit, apply the energy principle. Dene the system
as the HST
and
Earth. Treat both Earth and the HST as point particles, meaning that they only have rest
energy and kinetic energy.
And of course, there is energy due to their gravitational interaction, which is
called gravitational potential energy. The energy principle states that
∆Esys
Ef − Ei
Ef
= Wsurr
= W
= Ei + W
Substitute the energy of the system which is the sum of the particle energies and their interaction energy.
Ef
EEarth,f + EHST,f + Ugrav,Earth,HST,f
= Ei + W
= EEarth,i + EHST,i + Ugrav,Earth,HST,i + W
Write the particle energy as the sum of its rest energy and kinetic energy. Note that rest energy cancels out
of both sides of the equation, if we assume that the mass of the HST and Earth haven't changed during the
process.
Erest,Earth,f + Erest,HST,f + Kf + Ugrav,Earth,HST,f =
Erest,Earth,i + Erest,HST,i + Ki + Ugrav,Earth,HST,i + W
( E (((
(+ K + U
Erest,Earth,f
(
(((( + (
rest,HST,f
f
grav,Earth,HST,f =
(
( + E + K + U
(
(
(
E
(
i
grav,Earth,HST,i + W
(rest,Earth,i rest,HST,i
Kf + Ugrav,Earth,HST,f = Ki + Ugrav,Earth,HST,i + W
Solve for the work done on the system by the surroundings.
W
= Kf + Ugrav,Earth,HST,f − Ki − Ugrav,Earth,HST,i
Substitute expressions for kinetic energy and gravitational potential energy.
W
=
−GmHST MEarth
1
−GmHST MEarth
1
mHST vf2 +
− mHST vi2 −
2
rf
2
ri
Now, we need to determine values for initial and nal speed and initial and nal distance from Earth.
Initially, the HST is at rest on Earth, but Earth is rotating with a period of 24 hours. The average radius
of Earth and HST's initial distance from the center of Earth is
of the HST is
ri = 6.371 × 106 m.
Thus, the initial speed
v
2πR
T
2π(6.371 × 106 m)
=
(24 h)(3600 s/h)
= 463 m/s
=
The nal speed and nal distance of the HST in orbit was computed in part (a) as
rf = 6.427 × 106 m.
W
=
W
=
vf = 7871 m/s
and
Substitute all known values into our expression for work.
−GmHST MEarth
1
−GmHST MEarth
1
mHST vf2 +
− mHST vi2 −
2
rf
2
ri
W
−G(11000)5.97 × 1024
−G(11000)(5.97 × 1024 )
1
1
(11000)(7871)2 +
− (11000)(463)2 −
6
2
6.427 × 10 m
2
6.371 × 106
10
11
9
11
= 3.407 × 10 J − 6.815 × 10 J − 1.179 × 10 J + 6.875 × 10 J
W
=
3.9 × 1010 J
As expected, the work done on the system is positive. Because the HST in orbit is further away from Earth,
the system has a greater potential energy. Also, in orbit it is traveling faster (7900 m/s) than when on the
launch pad on Earth (460 m/s). Thus, the nal energy of the system is greater than the initial energy of
the system. As a result, the work done on the system should be positive, according to the energy principle.
Our result at least has the correct sign.