Gareth56

Thanks but I don't suppose you could expand your reply for layman

What the difference (physics wise) from the normal force when your standing still and when siting in a rollercoaster car.

At the top of a rollercoaster loop gravity is sufficient to provide the necessary centripetal force, however at the bottom of the loop gravity is pointing away from the centre so other than the normal force (which I presume balances mg) where does the extra inward force come from which supplements the normal force thus contributing to the feeling of heaviness and keeps you moving in a circle?

Thanks

I'm obliged. I thought I'd check with the experts before taking what I read at face value.

The reason I asked the question was I read the following quote on:-

Weightlessness

"The condition of zero-gravity (zero-g) or microgravity experienced by all spacecraft and their occupants when in orbit, in any other state of free fall, or traveling through space at constant speed."

If gravity has infinite range why do you feel weightless if you're traveling through space at constant speed? For example if I were traveling (in a space craft) at constant speed towards the Moon why would I feel weightless?

I understand there's now an 8th edition so hopefully the proof reader spotted the typo.

Here is the exact quote from the book:-

"Why is the launch area for the European Space

Agency in South America and not in Europe?

Explanation:-

Satellites are boosted into orbit on top

of rockets, which provide the large tangential speed

necessary to achieve orbit. Due to its rotation, the surface

of Earth is already traveling toward the east at a

tangential speed of nearly 1700 m/s at the equator.

This tangential speed is steadily reduced further

north, because the distance to the axis of rotation is

decreasing. It finally goes to zero at the North Pole.

Launching eastward from the equator gives the satellite

a starting initial tangential speed of 1700 m/s,

whereas a European launch provides roughly half that

speed (depending on the exact latitude)."

In College Physics 7th Ed by Serway it states that the tangential speed of the surface of the Earth at the equator is 1700m/s!!

Is this correct? I thought it was more like 465m/s.

With respect to observable I'm a tad unclear as to what is the un-observable universe and is it possible to determine it's size. For that matter what's the size of the observable universe?

Thanks.

Thanks for the replies. So the consensus of opinion is that the gravitational attraction exerted by these planets (Super- Jupiters) on the gas molecules/atoms is sufficiently great to prevent them from being blown off by the parent star's radiation.

If hot jupiters (e.g. 51 Peg b) orbit so close to their parent star why doesn't the heat or radiation or solar wind blast the gas off the hot jupiter?

Anything you like if you can explain it

I've tried distributing etc but cannot get the correct answer of 60.1N

My book doesn't give the steps.

Tsin37 + (1.33T)(sin53) - 100N =0

I never know what to do with the (1.33T)(sin53)

Any assistance would really be appreciated.

Sorry but I don't understand that.

I always thought from the "firing a canon ball fast enough off a mountain" experiment that if you were going fast enough you would always fall at the same rate as the curvature of the earth.

I'm not too sure about falling to the Moon either.

I understand that when one is orbiting the earth in say the Shuttle the feeling of weightlessness is due to you "falling" with the curvature of the earth, however if you are say half way to the Moon and traveling in a "straight line" the force of gravity due to the earth is about 1.3 (if I've worked it out correctly) so would you be floating around in your space capsule? 1.3 is only slightly less than the acceleration due to gravity on the Moon which you can walk on?

Thanks for the reply. What puzzles me is why the mg appears outside the bracket? and there's only a 'g' in one of the terms so how can you distribute 'g'?

Apparently mg + mv^2/r = mg(1 + v^2/rg) but does anyone know what is the name of the process?

Firstly Happy New Year to all forum users.

I've just watched the film 'Superman Returns' (for the physics mistakes of course!) which was on over the holidays and noticed that when Superman carried the "kryptonite island" off into he space he gave it an almighty push and off went the island into space however Superman stayed where he was. Wouldn't Superman have been propelled backward with an equal force to that which he gave the island, or are we to assume that Superman has infinite inertia?

Thank you.

All was going swimmingly well reading my College Physics book by Serway until the dreaded simultaneous equation which I hope somebody can help me with. I can handling them except when there are letters involved.

Find 'd'

I managed to get to this point then collapsed! Units have been omitted for clarity.

-d sin35deg = -0.5(9.81)((d cos35deg)/25)^2

Apparently d = 108m

Thanks

Think about it. If you hung a bunch of gold from a spring scale on the moon (where gravity is weaker and doesn't pull as hard), would you need more or less gold to make the scale read "1N" compared to being on the moon?

Using the equation weight(N) = mass x accel. due to gravity yields the following (I hope)

Moon

1N/1.6m/s^2 = 0.625kg

Earth

1N/9.81m/s^2 = 0.102kg

So it looks like to requires more gold on the Moon!

Thanks:-)

Is a 1 Newton of gold worth more on the Moon than on Earth?

Thanks. I forgot that little trick!

Could some kind soul explain the simplification of the following

sin(theta) + mucos(theta) =

cos(theta) - musin(theta)

tan(theta) + mu

1 - mutan(theta)

I understand that sin(theta)/cos(theta) = tan(theta), but cannot work out how the 1 - the rest comes from.

Thanks

I love these forums

I love these forums

If the maths isn't too complicated or the explanation not too involved what does this mean?

"It can actually be thought of as a negative force, in essence, pulling water away from the Moon and away from Earth's surface -- a second high tide.