Cars go loopy
Question: I have just seen a TV advert in which a car travels along a
straight track which then veers upwards to loop the loop before returning to
ground level. The car follows the road without falling off, even when it is
upside down at the top of the loop. Special effects aside, can a car loop the
loop, and, if so, how fast would it have to travel?
Answer: Surprisingly, the minimum speed a car must be travelling at to loop
the loop depends only on the force of gravity and radius of the loop. The speed
required does not vary with the mass of the car. This is because to loop the
loop the force flinging the car outwards (centrifugal force) must equal the
weight of the car at the top of the loop. The equation for both forces is the
same (mass 脳 acceleration), so the mass can be cancelled from the
calculation.
Many a schoolchild is puzzled by the difference between weight and mass and
this problem highlights the difference. The car鈥檚 weight is the force pulling it
down and is given by mass multiplied by acceleration due to the gravitational
attraction of the Earth. For a car of mass M kilograms this is
M 脳 9.81 newtons.
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To counter this the car must be travelling fast enough for the centrifugal
force to equal this at the top of the loop. If the radius of the loop is
r metres and the velocity of the car is vmetres per second, the
formula for this is M 脳 v+2/r. Equating these two opposing
forces and cancelling M gives 9.81 = v2/r,
remembering that velocity is in metres per second and radius in metres.
So for a 10-metre radius the car must be travelling at the square root of
98.1 metres per second, or about 10 metres per second. That鈥檚 36 kilometres per
hour.
This is surprisingly low, but remember it鈥檚 the speed required at the top of
the loop after climbing, so a higher speed would be required at the bottom. An
obvious question leading on from this is from what height would a car have to
freewheel to loop the loop?
This time the answer does not even depend on gravity. The above equations
give v2 = r 脳 9.81 for the minimum speed at the top of the
loop. Equating potential energy gained by falling from height h metres
with kinetic energy gained at velocity v gives 0.5 M 脳 v+2 =
M 脳 9.81 脳 h or v+2 =
2 脳 9.81 脳 h. Combining the two
equations gives r 脳 9.81 = 2 脳 9.81 脳 h or h =
r/2
In other words, dropping from half a radius higher than the top of the loop
is sufficient (though friction losses would in practice necessitate a greater
height). This seems remarkably small and I did not believe it, until trying it
with a marble and flexible curtain track made it seem plausible.
Colin Kerr
Edinburgh
New star
Question: How large will the International Space Station have to be before it
can be seen with the naked eye? And when built, how significant will it appear
to an observer on Earth?
Answer: A satellite does not need to be very large to be visible to the naked
eye. On any dark night you can see several overhead. In fact, under the right
conditions, both the Mir space station and the space shuttle can be as bright as
the brightest stars.
What determines the visibility of a satellite is not its size, but other
conditions. Although the sky must be quite dark, the Sun must still be shining
on the satellite to illuminate it. For high-altitude satellites this condition
is common because they orbit outside the shadow of the Earth.
However, for the shuttle and any space station which orbits at less than 500
kilometres above the Earth, the time when all these conditions are met is
relatively brief. Astronomical websites frequently publish times when major
satellites are visible, including the shuttle and Mir.
Barry Spletzer
Albuquerque, New Mexico
Answer: The International Space Station can already be seen with the naked
eye. In the early evening or early morning, about one hour before sunrise or
after sunset, lots of satellites can be seen as they reflect the Sun鈥檚 rays. The
German Space Operations Centre has a great Web page at
http://www.gsoc.dlr.de/satvis/
which gives predictions for visible satellites anywhere in
the world. However, I haven鈥檛 spotted a satellite while using its predictions
yet, because it has been cloudy or raining every single time I have tried.
Alan Garde
North Lambton, New South Wales
Answer: Wait no longer! The station is already visible, and was from the time
that the first module, Zarya, was launched. It currently appears as a bright
鈥渟tar鈥� of about 0 magnitude鈥攕lightly less bright than Mir, which is also
regularly visible. When completed, the ISS is likely to rival Venus for
brightness at about magnitude -4.
To discover when the station is visible from your location you need to enlist
the help of NASA via the Marshall Space Flight Center by accessing its excellent
website at http://liftoff.msfc.nasa.gov/realtime.
This will show you the current orbit and position of the station and will
also allow you to check pass times for your latitude and longitude. As an
alternative I would recommend using the equally excellent STS-Orbit Plus program
available as shareware at http://tie.jpl.nasa.gov/dransom/stsplus.html.
You will also need regularly updated orbit parameters (or elements), which are
available from a number of sources, including
http://celestrak.com/NORAD/elements/index.html.
The STS-Orbit Plus program will run on all PCs using DOS or Windows (all
versions) and will display, on your home computer, satellite tracks of the type
seen on television from NASA Mission Control. Times of visibility will be
correct to within seconds as long as the orbital elements are up to date.
Simon Warner
Middlewich, Cheshire
This week鈥檚 question
Touchy taps: I have a shower hose that attaches to the hot and cold taps in
my bath. When I want to use the shower I turn on the hot tap fully and then,
very slowly, turn on the cold tap in the belief that the water temperature will
adjust slowly until it is just how I want it.
However, the temperature instead turns dramatically from scalding hot to icy
cold even though the adjustment is slow. Why is this and why
does the temperature not change gradually?
Edwin Colyer
Manchester