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Physics, help me please.

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andrew.
Posts: 2349 - Karma: 165
hey, so my physics teacher gave us an assessment today, basically everyone got 2 different questions, and we have to solve them and explain our work, and im having trouble with them. i missed a good chunck of one section, and of coure, thats the section my questions came from. im not asking anyone to do my work, but if someone who gets physics is bored, could you maybe help me out by doing it out and comparing answers? anyone who helps gets a hefty 10/10.

1. Determine the minimun angle at which a roadbes should be banked so that a car traveling at 10.0 m/s can safelt negotiate the curve if the radius of the curve is 2.00 x 10^2 m.

2.What is the acceleration due to gravity at an altitude of 1.00 x 10^6 m above the earth's surface? NOTE: the radius of the earth is 6.83 x 10^6 m.

i had some trouble with number one because i never took a calc class, so im stuck with tan(theta) on one side and i dont know what to do. any help at all is greatly appreciated.
Dec 4 2008 3:40PM
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GreekAR6
Posts: 527 - Karma: 11
I have no idea for the first one, but isn't the second one always 9.8 meters per second squared? (if air resistance is negated, which I'm assuming)
Dec 4 2008 3:57PM
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andrew.
Posts: 2349 - Karma: 165
thanks for the reply, but theres some wierd formula to use when it gets that far away from earth. something like the centripetal acceleration is the gravitational pull, or maybe its the centripetal force? either way, its not 9.8. 10/10 either way tho.
Dec 4 2008 4:02PM
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Microcosm
Posts: 4765 - Karma: 153
nah, it changes ever so slightly depending on your distance from the earth. 9.8m/s is generally used as it's not necessary to determine the EXACT acceleration and any variations are pretty infinitesimal. his question is probably asking for the EXACT though as more of concept.

where are you stuck on 1? I don't feel like thinking about it conceptuallyl, but I could try the math.
Dec 4 2008 4:05PM
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andrew.
Posts: 2349 - Karma: 165
the math equation is 20=squareroot(200*9.8*tan(theta))

sorry if that doesnt make much sense, but my computer doesnt have theta or square root. idk if its right tho, because i get a negative answer when i do it out.
Dec 4 2008 4:08PM
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Fly_White_Boy
Posts: 301 - Karma: 13
Second question:

(gravity on earth/gravity at given altitude)=(radius at given altitude/radius of earth)^2

so if you plug in your numbers, you get:

(9.8/x)=(7,830,000/6,830,000)^2
(9.8/x)=(1.146)^2
(9.8/x)=1.313
9.8=1.313x
7.464=x=gravity at given altitude

So, at 1.0 x 10^6 m off Earth, the force of gravity would be 7.464 m/s^2.

Dec 4 2008 4:16PM
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Microcosm
Posts: 4765 - Karma: 153
20=√(200*9.8*tan(θ))
20^2=1960*tan(θ)
400=1960*tan(θ)
400/1960=tan(θ)
tan^-1(400/1960)=θ
θ=11.5346°
Dec 4 2008 4:19PM
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Fly_White_Boy
Posts: 301 - Karma: 13
Your calculator might be on radian mode if its a graphing calculator, try setting it to degree mode.
Dec 4 2008 4:19PM
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andrew.
Posts: 2349 - Karma: 165
no homo, i love you both. i suck so bad at calc its not even funny. the only problem i have is that i dont think we learned the equation for the acceleration one yet. ill keep looking for it, but i hvent heard of it. what is it in the original form?
Dec 4 2008 4:24PM
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eazylax
Posts: 206 - Karma: 17
2V=√(r*a*tan(θ))

a= acceleration due to gravity in this case
Dec 4 2008 4:39PM
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Fly_White_Boy
Posts: 301 - Karma: 13
You just put Earth's gravity (9.8 m/s^2), your altitude (plus the radius of the earth, which gives you the height above the center of the earth), and the radius of the earth in, and solve for the gravity at that altitude, which i made x in that equation. you know, just do out the algebra
Dec 4 2008 4:41PM
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bbtdeathcross
Posts: 59 - Karma: 26

1)

for the first one you use f=ma, where a=v squared/r. so you want to find the angle where that will push inward just as hard as the car is accelerating outward. so sin(theta)*m=m(v squared/r). The ms cancel out, so sin(theta)=v sqaured/r. On your calcuator you can do inverse sin of v squared/r.

2)

f=ma, a=f/m.

I think you use the force eqn f=gm/r or something like that.

Sorry if these arent right, im a little rusty.

Dec 4 2008 4:56PM
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andrew.
Posts: 2349 - Karma: 165
^its cool, help is help. i used v^2=square root(rg tan theta) and got a reasonable answer (well i didnt, it was someone above), but ill plug everything into f=ma and see what i get. as for number 2, only problem with that is we dont have a mass or force. we get radius of earth and altitude.
Dec 4 2008 5:01PM
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killa_cam
Posts: 13143 - Karma: 15
For the second one, use the universal gravity formula.
F=Gm1m2/r^2

F = ma, so divide by mass m2

a = Gm1/r^2

where G is the universal gravitational constant, m1 is the mass of the Earth, and r is the distance from the centre of the Earth (altitude + diameter of Earth(r/2))
Dec 4 2008 5:09PM
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