Download 88 Hubble.p65

Physics Factsheet
www.curriculum-press.co.uk
Number 88
The Physics of the Hubble Telescope
Power
The Hubble telescope is a technological masterpiece. It has been
reported to be 50 000 times more sensitive than ground-based
telescopes, and has 10 times better resolution.
The instruments and tracking systems of the telescope need power.
This is provided by two large solar panels, each about 2m by 12m.
They produce about 2800 watts of electricity, which is stored in 6
batteries aboard the spacecraft. Each battery can provide enough
energy to keep the telescope and instruments in normal operation
for over 5 hours.
We will to look at certain specific aspects of the Hubble telescope,
and see how they link to Physics at A-level.
The orbit
Many people are under the illusion that the Hubble telescope is in
a geostationary orbit, like many communications satellites.
Problem 1 : What are the advantages of a geostationary orbit?
Answer:
(i) the satellite will always ‘observe’ the same area on the
Earth’s surface.
(ii)the satellite can be in constant communication with the
same point on the Earth’s surface.
solar pannels
These are useful for a communications satellite, but not necessarily
for the Hubble telescope. For a start, the telescope observes objects
in space, and not on the Earth. There is no need for it to be in
constant communication with any specific point on the Earth’s
surface.
Problem 2: Suggest two reasons why the telescope is not
operated directly from the solar panels.
Answer:
(i) Certain operations might require power greater than 2800
watts for a brief period of time.
(ii)The solar panels are in the Earth’s shadow for part of each
orbit.
The most important factor is that the telescope is above virtually all
of the Earth’s atmosphere. As the atmosphere is already significantly
thinning by the time you reach the top of Mount Everest, there is
no need to be in a very high orbit.
Steering
As the telescope is in a circular orbit around the Earth, it must
continually keep adjusting its orientation to keep itself fixed onto a
target in space. Some observations can take hours.
Secondly, the telescope was launched, and was regularly serviced,
by the Space Shuttle programme. Critical flaws in the original optics
were corrected by Shuttle missions.
orbit
The Hubble telescope is in a low Earth orbit, about 600km above the
Earth’s surface. Compare this to a geostationary orbit at an altitude
of 36000km.
600km
A = Hubble
B = geostationary satelite
Earth
6400km
A
There are three parts to the steering system.
(a) spinning gyroscopes keep track of the general orientation of
the telescope in space.
(b) fine guidance sensors lock onto ‘guide stars’ in roughly the
same direction as the target.
(c) the telescope is then rotated by reaction wheels. These are like
flywheels, and are spun by electric motors controlled by the
spacecraft’s computers.
B
36 000km
Exam Hint: Make sure you are confident with circular motion
calculations. There will always be questions on forces, speeds,
radii, etc. See question 1 at the end of this Factsheet.
These reaction wheels are an example of Newton’s Third Law – for
every action there is an equal and opposite reaction.
1
Physics Factsheet
88. The Physics of the Hubble Telescope
Black holes
If a reaction wheel is spun clockwise, it will exert an equal force on
the spacecraft causing it to rotate anticlockwise. By motion in the
three perpendicular directions, the spacecraft can be shifted into
any orientation.
The Hubble telescope can by used to locate black holes. Because
of the immense mass at its centre, light cannot escape from the hole
itself. But material spinning around the outside of the black hole
does emit radiation which we can observe.
Reaction
Because of the extreme gravity present, this material must orbit the
black hole at very high speed, giving rise to a measurable blue shift
at one side of the hole, and red shift at the other.
Action
Black hole
Using these three techniques, the telescope can remain locked onto
a target over a whole orbit (or longer).
Problem 3: Should these reaction wheels be light or heavy?
Red shift
Blue shift
Answer: Heavy wheels would provide a greater reaction force
as they spin, but they would be more difficult to control for
fine adjustments to the position of the craft.
Observer
Optics
There are a number of telescopes and related optical devices on the
spacecraft. The detail of them is beyond the remit of this Factsheet,
but together they enable measurements over a wide range of
wavelengths from infra-red through the visible spectrum to ultraviolet.
In one black hole observed by the Hubble telescope, this material is
spinning around the black hole at an orbital speed of 400kms-1.
Calculations predict the black hole to have a mass of over three
hundred million times that of the Sun.
Sensitivity and Resolution:
Problem 4 : Why not just observe in the visible region?
At the beginning we mentioned the sensitivity and resolution of
the Hubble telescope. Make sure you know the difference between
these terms.
Answer: The greater the range of wavelengths observable,
the more information can be obtained. Different objects in
space emit radiation in different regions of the e.m. spectrum,
right down to radio waves.
Sensitivity is a measure of how responsive to low light levels the
telescope is; resolution is concerned with the ability of the telescope
to distinguish between two point sources (e.g. stars) that are
separated by only a small angle in the sky.
In addition, dust clouds can absorb visible light from sources, but
often not infra-red. Some stars within or beyond a dust cloud can
only be observed by their infra-red emissions.
An example of what a telescope might make of two such light sources:
IR only
visible
plus IR
Dust
Problem 5: The infra-red and visible emissions of stars A and
B are separately observed (by means of suitable filters).
A
B
A
B
IR
Visible
High sensitivity
low resolution
A bright blur
Low sensitivity
high resolution
Two very dim
points
High sensitivity
high resolution
Two very bright
points
Which star is hotter?
Answer: Star B is at a higher temperature. A greater proportion
of its energy is being emitted at shorter wavelength / higher
frequency.
Exam Hint: Learn definitions precisely. A vague definition of
‘getting a better image’ would be of little value.
2
Physics Factsheet
88. The Physics of the Hubble Telescope
Questions
6. The Hubble telescope cannot study Mercury or Venus properly.
They are aligned too near the Sun. The intense radiation from
the Sun would damage the sensors and sensitive electronics in
the instruments aboard the spacecraft.
1. (a) What are the equations for gravitational force and centripetal
force?
(b) Equate these expressions, and rearrange to find an expression
for orbital speed of a satellite.
5. See the text for the explanation. In practise, each has its own
use. High sensitivity lets us ‘see’ further into space. High
resolution lets us separate (and study) individual light sources
in space.
(c) Given the values G = 6.67 × 10 -11 Nm 2 kg -2 , and
M = 6.0 × 10 24kg, find the orbital speed of the Hubble
telescope. (M is the mass of the Earth)
Observed λ = (5.000 × 107 ) + (6.67 × 10-10 ) = 5.007 × 10-7m.
4 × 105
× 5.000 × 107 = 6.67 × 10-10m
3 × 108
2. From the data given in the ‘Power’ section, work out:
(a) the total area of the solar panels.
4. ∆λ =
(d) What is the orbital period (time for one revolution) of the
Hubble telescope?
3. The exhaust gases would affect the sensitivity and resolution
of the telescope, just as the Earth’s atmosphere does.
(b) the power generated by each square metre of a panel.
(c) Incident power
= 58.3 × 2 = 117W.
m2
(c) assuming the panels are 50% efficient, what is the energy of
the solar radiation striking each square metre of the panels?
(b) Power
2800
=
= 58.3W.
48
m2
3. Suggest why small rockets could not be used to adjust the
orientation of the telescope.
2. (a) A = 2 × 2 × 12 = 48m2.
=
2πr
2π × 7 × 106
= 5787s, or 96 min.
=
v
7600
∆λ
∆λ
v
=
=
λ
c
f
(d) T = distance
speed
4. The red or blue shift due to relative motion of celestial objects is
given by:
(c) v = 7600ms-1, or 28 000kmh-1 (r = 6400 + 600 = 7000km )
If the material mentioned (about the black hole) is moving away
from us at a speed of 400kms-1, find the observed wavelength of
light emitted at an original wavelength of 5.000 × 10-7m.
GMm , becomes v = v(GM / r)
(b) mv2
=
r
r2
FC = mv2
r
5. Explain the difference between resolution and sensitivity of the
Hubble telescope. Say which you think is more important.
Answers
1. (a) FG = GMm
r2
6. Suggest which two planets the Hubble telescope cannot
properly study. Give a reason.
Acknowledgements:
This Physics Factsheet was researched and written by Paul Freeman
The Curriculum Press,Bank House, 105 King Street,Wellington, Shropshire, TF1 1NU
Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber.
No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher.
ISSN 1351-5136
3