Download 103 R and AW.p65

Physics Factsheet
www.curriculum-press.co.uk
Number 103
Real And Apparent Weightlessness
When you are standing on the ground, you feel comfortable because there are 2 equal and opposite forces acting on you – your WEIGHT
(the gravitational attraction of the Earth pulling you down) and the REACTION of the floor pushing up on you. [Which one of Newton’s
Laws applies here??*].
The one that you are actually aware of is the REACTION force, and when that is equal to your weight you feel normal. But if the reaction
force changes, then you feel heavier or lighter than usual.
In this factsheet you will recall some of the situations in which you feel weightless. Either the WEIGHT disappears altogether, or it is being
used to do something else and the REACTION disappears, so you float off the ground.
Geoffrey
Giles
Weight = mg
Reaction
Remember this from GCSE??
Your MASS is the amount of matter inside you. It is
measured in kg and is the same everywhere in the
Universe (unless you lose an arm or leg on your way
to the Moon).
• Your WEIGHT is the force of gravity due to the Earth
attracting you towards its centre. Weight is a vector
and is measured in Newtons. It gets less as you move
away from the Earth and is different on other planets.
• If you know the MASS, multiply by 10 to get the
WEIGHT (on Earth). The Earth’s gravitational fieldstrength g is 10 N/kg or 10 m/s2.
• Any two masses attract one another, but the only one
that usually affects you is the EARTH, because its
mass is so enormous compared to all the other masses
around.
…or unless you are travelling at nearly the speed of
light (A-level exception to the rule).
•
This means it can be represented by an arrow of the
right length and its direction makes a difference.
At AS-level, we need to be more precise. g is really 9.8 or
9.81 (check your own specification for their requirements).
And the units must be given as N kg-1 or ms -2.
Can you estimate your own weight in Newtons (on Earth)? If you keep this figure in mind, it will help you to avoid mistakes and have a
clearer idea about what happens when your weight changes. A-level boys probably weigh about 800 N, and girls about 700N.
* It’s Newton’s First Law. They are two different types of force acting on the same object, so they can cancel out and you are in
equilibrium. Equilibrium feels comfortable!
1
Physics Factsheet
103. Real And Apparent Weightlessness
(1) In Space
Steady speeds in space
There really IS no weight because you are so far from any planet or
star that there is nothing to exert a force of attraction upon you.
A spaceship far away from any planet or star will experience NO
forces, neither a pull of gravity towards something nor any air
resistance opposing the motion. So if it is hurtling along at 50 km s-1,
it will just keep going at that speed in a straight line because there
is no unbalanced force to make it accelerate or slow down. (Which
one of Newton’s Laws applies here?*)
Common pupil mistake: Some people think that because the
atmosphere is a thin layer around the Earth, and there’s no air in
space, the same applies to the force of gravity. But gravity gets
less gradually as you move away from the Earth, and is still a
sizeable force far above the top of the atmosphere. The radius of
the Earth is 6400 km (6.4 × 106 m). If you were 10× this distance
away from the Earth’s centre, your weight would be one
hundredth of its usual value. About 7-8 Newtons, the same as a
bag of apples (use your estimated weight to check this value).
Do you think this is low enough to count as zero? Probably not.
Try 10× further again. You would still be nowhere near Venus or
Mars to be affected by their gravitational pull on you. Look at
the graph below to see how the force of gravity on you changes
with distance from the Earth.
Earth
If you are inside that spaceship, you will be hurtling along with it at
the same speed, and genuinely weightless. You will be quite
comfortable, but floating about inside it. You would have to be
strapped down to a chair or bed, as there’s no reaction force on you
from anything.
uniform velocity 50 km s-1
(this means the same as steady
speed in a straight line)
Gravitational field lines show the
direction of the force of attraction on
a mass at any point.
Accelerating in space
If you want to speed up, slow down or change direction in space,
you have to produce your own force. The spaceship carries small
rocket engines, usually turned off. To speed up, the rocket produces
a jet of gas backwards.
(NB This is a gravitational field, NOT
a magnetic field).
The rocket produces a backwards force on the jet, so the jet
produces a forwards force on the rocket (attached to the spaceship)
and it will accelerate. (Which one of Newton’s Laws applies this
time?**)
Change in your weight as you move away from the Earth
800 N
700 N
600 N
Backwards force on
gas
500 N
Forwards force on
space ship F
400 N
300 N
No air resistance or drag
200 N
100 N
How about you?
0N
0
1
2
3
4
5
6
7
8
Distance from centre of Earth in multiples of Earth radius
Inside the ship, you will carry on moving at the original speed, until
the back of the spaceship catches up with you. Then there will be
a reaction force acting forwards on you, and you will accelerate as
well. If the spaceship accelerates at 9.81 ms-2, you will be standing
on the back of the spaceship and experience just the same reaction
force as you do on Earth. The resultant force on the space ship will
be the same as F, since there’s no air resistance to act against it.
The resultant force on you will be the reaction force.
How would you describe the shape of this graph? What
relationship does it represent? [4]
Answer from a weak student:
Your weight gets less as the distance from the Earth increases[1].
It does not decrease evenly, but more steeply when you are closer
to the Earth[1].
The resultant is usually the single force that combines
the effects of several forces acting on the same object. But
when there’s only one force acting, the resultant is the same as
that one force.
That’s a good GCSE answer, but worth only 2 marks at A-level.
It does not answer the second part of the question.
What the examiner would be looking for
* Newton’s First Law, again. Hopefully you know it off by heart.
** Newton’s Third Law. Two different objects produce equal and
opposite forces of the same type on each other.
The weight decreases as the distance from the Earth increases[1].
When the distance from the centre doubles (e.g. from 1 Earth radius
to 2), the weight decreases from 800 N to 200 N[1], so becomes 4
times less.[1] This is an inverse square relationship[1].
At A-level, you need to use the numbers on the graph and use the
correct description for the relationship. “Inverse” means that
the weight gets smaller, and “square” means that you divide by
22. Check that this also works going from 2 radii to 4.
2
Physics Factsheet
103. Real And Apparent Weightlessness
So using Newton’s Second Law (in its familiar form, F = m × a)
Worked example
(a) If g = 9.81 ms-2 on Earth, and the radius of the Earth is
6.4 × 106 m, find the value of g (g’) at a height of 12.8 × 106 m
above the Earth’s surface.
(b) A satellite is orbiting at a height of 12.8 × 106 m from the
surface. An astronaut weighs 1000N on Earth. Find his
weight when in the satellite.
19.2 × 106m
mg’ (Force) = m (mass) × v2/r (acceleration)
The m’s cancel, leaving g’ = v2/r
You may not like algebra, but if you think about the meaning of
these 3 quantities you will see a very important result. If we put a
satellite up at a particular height, there is only one possible value of
g’, (since everything accelerates at 9.81 ms-2 on Earth, and you
calculate g’ from 9.81 and the height).
Hints. You will come to grief in
this type of question if you don’t
draw a diagram.
Distances are always measured
from the centre of the Earth, and
if you forget this you will get the
wrong answer.
So having decided on the value of r, there is only one possible
value for v. All satellites at that height have to be travelling at
exactly that speed, chasing each other around in a circle. The mass
is unimportant; it cancelled out.
If they go too fast, they leave the orbit and go off into outer space,
and if they go too slowly they spiral inwards and crash into the
Earth.
(a) Total distance from the centre of the Earth
= (6.4 + 12.8) × 106 m = 19.2 × 106 m.
This is (19.2 × 106) ÷ (6.40 × 106) = 3 times further away
from the centre than usual.
Gravitational forces obey an inverse square law, so divide
9.81 by 32.
g’ at this height = 9.81/9 = 1.09 m s-2 (One-ninth of the value
on Earth)
(b) If he weighs 1000 N on Earth, his mass is 1000 ÷ 9.81 kg =
101.9 kg, which does not change. On the satellite his weight
will be mg’ = 101.9 × 1.09 = 111 N.
This is also ONE NINTH of his weight at the surface.
So when you are in orbit, you have to be travelling at precisely the
right speed and in the right direction, otherwise you would not stay
in orbit. The gravitational force acting on you is precisely the
correct force to supply the centripetal acceleration, and with no
other forces acting on you, you will continue to travel in that circle.
You will feel completely weightless because you will float around in
the satellite and there will be no reaction forces on you from any of
the surfaces around you, exactly as in zero gravity.
Worked example. Do you like high-speed travel?
How fast are these satellites actually going?
We can easily find the speed of a satellite at a height above the
surface of 12.4 × 106 m, since we know g’ already.
(2) In Orbit
When in orbit, you experience all the effects of weightlessness
even though the weight is not zero. This is because the force of
gravity acting on you is exactly the right strength for maintaining
you in orbit and doing nothing else. You are not pressing on the
floor or walls of the space craft so they are not pressing on you –
there are NO reaction forces. You are travelling in a circular path
and because velocity is a vector and its direction is changing, you
are in fact accelerating towards the centre of the circle even though
you do not feel as if you are. (See factsheet 19 on circular motion).
g’ = v2/r can be rearranged to give
v2 = r × g’
v2 = (19.2 × 106) × 1.09 = 2.09 × 107
so v = 4.57 × 103 m s-1, or 4.57 km s-1.
Since time = distance / speed, we can find how long it would
take to orbit the Earth.
Time
velocity v
Gravitational force
of attraction
2πr
(4.57 × 103)
2 π x (19.2 × 106)
=
(4.57 × 103)
= 2.64 × 104 seconds.
2.64 × 104
or
= 7.3 hours.
3600
=
The Moon takes 28 days to orbit the Earth so must be much further
away than that. Weather satellites take 1.5 hours to orbit the Earth,
so they are a lot faster than in the above example and must be a lot
closer to the Earth. If you were on a weather satellite going at these
colossal speeds, you would not notice the speed and would feel
completely normal (apart from being apparently weightless).
If the satellite has mass m, the force of attraction on it is mg’, where
g’ is less than 9.81 ms-2 and is found by a similar calculation to the
one above.
To remain in a circular orbit, the acceleration towards the centre of
the circle has to be v2/r, where v is the magnitude of the velocity
(speed), and r is the distance of the satellite from the centre of the
Earth.
3
Physics Factsheet
103. Real And Apparent Weightlessness
(3) Artificial Gravity
(4) Lifts And Roller-coasters
It is not good for the human body to be weightless for long periods,
as your bones and muscles will deteriorate. Artificial gravity can be
produced in a space ship or satellite by making it spin about its own
axis. A jet of gas from the rocket engines could make it start to spin,
and it would then continue spinning since there is no drag force to
slow it down, exactly like Newton’s First Law. Inside it, the
astronaut’s body is forced to move in a circle rather than a straight
line, so the outer skin of the space ship will exert a force on you
pushing you inwards to provide the centripetal acceleration. If you
walk around on the inside wall of the space craft, you will feel a
REACTION force which could be designed to be the same as that
on Earth.
On Earth, even though weight does not change, lifts and rollercoasters certainly make us feel as if it does.
Geoffrey
Giles
W
W
R
Force on
rocket
Acceleration = 5 ms-2
r
Force on
gas jet
Reaction
force
Acceleration = 9.8 ms-2
Worked Examples
Spin
clockwise
speed v
If Geoffrey and Giles both have mass 75 kg, their weight W
is 75 × 9.8 = 735 N.
(a) Giles’ lift is accelerating at 9.8 ms-2, so he will accelerate at exactly
the same rate as the lift. There will be no contact between his feet and
the floor, so R = 0. Giles feels completely weightless.
Rocket
engine
(b) Geoffrey’s lift is accelerating downwards at 5 ms-2. The resultant
force on Geoffrey must make him accelerate at the same rate.
Resultant force = mass x acceleration (Newton 2 again).
W–R =m×a
735 – R = 75 × 5
= 375 N.
So R = 735 – 375 = 360 N.
Geoffrey will feel as if he weighs only 360 N.
If the astronaut’s mass is m, the radius of the space ship is r, and
the speed at which the outer skin is moving is v, then the reaction
force on the feet must be mv2/r (mass × acceleration, see fact sheet
19 on circular motion). To make him/her feel normal, v2/r needs to be
equal to 9.81 ms-2, or roughly 10 ms-2 as on Earth.
Worked Examples
Roller-coasters have exactly the same effect as a lift falling down a
lift-shaft, except that you probably survive at the end. If the lift
accelerates upwards, R will be greater than W and you will feel
heavier than usual.
1. Suppose the radius of the space ship is 10 m, then for v2/r to
be 10 ms-2, v must be 10 ms-1. [v =√(rg”), where g” is the value
of g that you want].
2. If r = 25 m, how many rotations per minute must there be to
give g” = 10 ms-2?
v = √(25 × 10) = 15.8 m s-1.
In one minute, his feet travel 15.8 × 60 = 948 m,
so number of rotations
= 948 / 2πr = 948 / (2 × π × 25) = 6 rotations per minute.
(5) Hump-back Bridges And Astronaut Training
If you drive over a hump-back bridge too fast, you certainly feel as
if you might fly off the seat of the car. This sensation, as always,
means that the reaction force approaches zero. It is rather like being
in orbit. The closer you are to the Earth, the faster you have to
travel to achieve the same effect. The curve of the bridge helps you
become weightless more easily. (NB This is very bad for your car!!)
3. If a space ship of radius 20 m is rotating at 0.05 rotations per
second, (3 r.p.m) what value of g” would you experience?
The same equation occurs again: mg = mv 2/r, but this time,
g = 9.8 ms-2 because you are on the surface of the Earth. The mass
cancels so the critical speed is unaffected by the mass of the car.
Distance travelled per second =( 2 × π × 20) × 0.05 = 6.28 m s-1 = v
g” = v2/r = (6.282/20) = 1.97 ms-2.
Slightly low values of g” will do you no harm, but this is too low to
feel normal, and spinning this fast will make you feel sick. Astronauts
have discovered that their bodies can tolerate 2 revs per minute but
not more. Also you might experience weird effects if your head has
a different g” to your feet. At 2 r.p.m. what radius of space-ship
would you need to get g” = 9.81 ms-2?
speed v
r
If the radius of curvature of the bridge is 15.0 m, how fast does
the car have to be going to leave the surface of the road?
v = distance travelled per second = (2πr) × 2/60 = (2πr)/30
so v2/r = (4 × π2 × r2) / (302 × r) = 0.0439 × r, which must be equal to g”
r = 9.81 / 0.0439 = 223 metres. Quite a sizeable spaceship!!
As before, mg – R = mv2/r, and R will be zero when this happens.
g = v2/r
v2 = r × g = 15.0 × 9.8 = 147
v = 12.1 ms-1
This is 0.0121 × 3600 km h-1, or 43.6 km h-1, which is equal to
about 23 miles per hour. If you are learning to drive, watch out!
4
Physics Factsheet
103. Real And Apparent Weightlessness
To give astronauts an experience of zero g (apparent weightlessness), it is much easier to use an aeroplane than a car. The plane needs
to travel in a curved path with a radius less than that of the Earth, and conditions similar to being in orbit can be achieved quite easily.
v
Commercial aircraft travel at about 500-600 miles per hour, or
about 250 ms -1. So to achieve apparent weightlessness,
v2/r = g, or r = v2/g = (250)2 / 9.8 = 6.4 × 103 m, or 6.4 km. This
is quite a gentle curve - not too difficult for an experienced
pilot to achieve.
r
Worked Exam Questions
2. A car is driven over a hump-back bridge which follows part
of a vertical circle of radius 25.0 m. Here is a free-body diagram
for the car at the highest point of the bridge.
1. A space station orbit is 210 km above the surface of the
Earth, which has a radius of 6370 km. The space station
takes 91 minutes to orbit the Earth once.
Normal reaction R
r
Weight W
The mass of the car is 950 kg. Calculate the normal reaction
force R
(i) When the car is parked on the bridge
(ii) When the car is travelling at 8.0 ms-1 (4)
(a) Calculate the velocity of the space station. (3)
(b) Calculate the acceleration of the space station. (3)
(c) A box of mass 5.3 kg is inside the space station. What is the
size and direction of the resultant force acting on the box? (3)
When the car is driven above a certain critical speed over the
bridge, it will lose contact with the ground and “take off”.
(iii)Explain why this happens.(1)
(iv)Calculate this critical speed. (2)
(v) Explain what is meant by “apparent weightlessness”, with
reference to this situation.(2)
Answers
(a) r = (210 + 6370) km = 6.58 × 106 m
2πr
velocity =
= 7.6 × 103 ms-1
(91 × 60)
(b) acceleration = v2/r = 8.8 m s-2 = g’
(c) resultant force = m × g’ = 5.3 × 8.8= 46 N towards the
centre of the Earth.
Answers
(i) When the car is parked, R = W by Newton’s first law.
R = mg = 950 × 9.8 = 9300 N (2.s.f.)
Exam Hint: You must not include more significant figures in
the answers than are given in the data provided in the
question. If you do not round off the answer to (a), your
answer to (b) will be slightly different: 8.7 ms-2. The examiner
will accept either 8.7 or 8.8 provided that you make it clear
what you are doing. The answer to (b) should be a bit less
than 9.8 ms-2 so this result should reassure you that you are
doing the question correctly. Section (c) is not asking for a
complicated calculation, just to check that you realise you
have just found g’. Note that the box will also be in orbit; i.e.
apparently weightless, but the resultant force on it is still mg’.
(ii) At 8.0 m s-1, centripetal acceleration
v2
(8.0)2
= =
= 2.56 m s-2.
r
25.0
This is provided by the resultant force towards the centre of
the circle, W – R.
So 9300 – R = 950 × 2.56 [ F = m x a]
R = 9300 – 2432
= 6900 N (2 s.f.)
(iii) It loses contact with the ground because the weight becomes
equal to the centripetal force and the reaction is then zero.
mass × v2
r
9300 × 25.0
2
v =
(950)
= 245 (ms-1)2
v = 15.6 m s-1
(iv) If 9300 =
(v) Apparent weightlessness occurs when you experience zero
reaction force from the ground, which occurs when the weight is
exactly right to provide the centripetal acceleration. At 15.6 m s-1,
the car will leave the surface of the road as if in orbit and the
passengers will also leave their seats. You feel as if W = 0 as
well as R.
5
103. Real And Apparent Weightlessness
Questions
Answers
Take the value of g on the Earth’s surface as 9.81 N kg-1, and the
radius of the Earth as 6400 km.
1. (a) The relevant base units are metre, kg and seconds. Apply the
formula Force = mass × acceleration to the N and multiply by
kg-1. You will get kg m s-2 × kg-1, which simplifies to m s-2.
(b) 608 N
(c) 91.7 kg.
1. (a) The units of g can be given as m s-2 or N kg-1. Reduce N kg-1
to base units and show that these are the same as m s-2.
(b) Find the weight of a girl of mass 62 kg.
(c) Find the mass of a man of weight 900 N.
2. (a) The 900 N man is on a satellite at a distance of 25,600 km from
the Earth’s surface. How many Earth radii is that measured
from the centre of the Earth?
(b) How much smaller will his weight have become at this
distance? (Hint: Use the inverse square law).
(c) Find the new value of his weight.
(d) What would the acceleration due to gravity be at this point?
2. (a)
(b)
(c)
(d)
5
25
36 N
0.392 m s-2.
3. (a)
(b)
(c)
(d)
300 N
300 N
0N
589 N
4. (a) 3540 m s-1. (3 s.f.)
(b) 15.8 hours
(Hint: Since the answer is given, you are expected to show
all your working and give an extra significant figure to prove
that you can do the calculation properly).
(c) 27.4 N (m × g’)
3. (a) If you are in a space ship accelerating at 5 m s-2 in a region of
space very far from any planet or star, and your mass is 60
kg, what will be the reaction force acting on you from the
back of the ship?
(b) How heavy will you feel?
(c) The space ship suddenly turns off its rocket engine so that it
stops accelerating and travels at a steady speed of 100 km s-1.
How heavy will you feel now?
(d) On Earth, what is your weight?
5. (a) v = 9 m s-1.
(b) Maximum v for comfort = 6.36 m s-1.
(c) No, they will not feel comfortable. They need to increase the
radius of their space ship so that it can rotate more slowly
and still give the correct velocity at the outer surface. [The
correct radius is 40.5 m, but that is a hard calculation. Follow
the one on the fact sheet if you want to try it].
4. (a) If a satellite is in orbit at a height of 25600 km from the Earth’s
surface, how fast is it going? (Hint: Use the value of g’ from
question 2 part (d), and don’t forget to convert r to metres).
(b) Show that it takes about 16 hours to orbit the Earth.
(c) What is the resultant force on a 70 kg astronaut inside the
satellite?
6. (a) 589 N
(b) 469 N
(c) 709 N
5. Some aliens from a planet where g = 4 ms-2 want to design a
spinning space ship to provide artificial gravity that makes them
feel normal. Their bodies can tolerate 3 rotations per minute.
(a) If the radius of the space ship is 20.25 m, find the velocity v
to give the right centripetal acceleration (value of g”).
(b) What is the maximum value of v their bodies can tolerate?
(Hint: Find the number of rotations per second and multiply
by 2 π r)
(c) Will they feel comfortable with this sized space ship? If not,
what improvement should they make in their design?
7. (a) 3930 N
(b) 8 N
(c) 70.04 m s-1!!
6. A girl of mass 60 kg is in a lift waiting to go down to the floor
below.
(a) What is the reaction force on her feet before the lift starts?
(b) What is the reaction force on her feet when the lift is
accelerating downwards at 2 m s-2?
(c) How heavy will she feel if the lift later accelerates upwards
at 2 m s-2?
7. (a) A racing-car of mass 800 kg travels at 70 m s-1 over a curved
bridge with radius of curvature 1000 m. What will be the
reaction force on the car at the highest point of the bridge?
(Hint: Use mg – R = mv2/r).
(b) What would be the reaction force if the radius of the bridge
was 500 m?
(c) How fast would the car have to be going to take off from the
bridge?
Acknowledgements:
This Physics Factsheet was researched and written by Hazel Lucas
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
6