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March 29, 2024, 07:59:21 am

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Rishi97

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Rishi's Physics Thread
« on: March 15, 2014, 02:28:41 pm »
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A 100kg meteor is falling towards the Earth from a distance of 4.0 Earth radii from the centre of the Earth (4.0Re)
a) Calculate g at this height
Pls provide a brief description of the formulas used
« Last Edit: March 15, 2014, 03:35:35 pm by Mr. T-Rav »
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Rishi97

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Re: Rishi's Physics Thread
« Reply #1 on: March 15, 2014, 04:31:41 pm »
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Explain why the gravitational field of the Earth does no work on a satellite in a stable circular orbit.

Thanks
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rhinwarr

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Re: Rishi's Physics Thread
« Reply #2 on: March 15, 2014, 04:58:17 pm »
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Quote
A 100kg meteor is falling towards the Earth from a distance of 4.0 Earth radii from the centre of the Earth (4.0Re)
a) Calculate g at this height
Pls provide a brief description of the formulas used


From the formula you can see the g is inversely proportional to the square of r.
If r increases by a factor of 4, g will be divided by 16
So g = 10/16 = 0.625 N/kg

Quote
Explain why the gravitational field of the Earth does no work on a satellite in a stable circular orbit.

Work is the energy applied to move an object in the direction of the force. The direction of the force acting on the satellite in orbit is the centripetal force which is towards the centre of the earth. However, the satellite does not get any closer to the Earth while it is in its circular orbit so its displacement in the direction of the force is zero. W=Fx so no work is done.

Rishi97

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Re: Rishi's Physics Thread
« Reply #3 on: March 15, 2014, 05:33:48 pm »
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OMG!!! Thank u soo much rhinwarr. You have no idea how long I've spent trying to figure this question out.
Thanks!!! :D
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Rishi97

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Re: Rishi's Physics Thread
« Reply #4 on: March 15, 2014, 05:47:43 pm »
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Neptune has a mass of 1.02 x 10^26 kg. Its moon Triton has a mass of 2.14 x 10^22 kg and an orbital radius of 3.55 x 10^8 m . For Triton, calculate it's:
a) orbital radius
b) orbital speed
c) orbital period
For part a, I'm unsure of what mass to use, Triton or Neptune?

Thanks ;)
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lzxnl

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Re: Rishi's Physics Thread
« Reply #5 on: March 15, 2014, 07:09:40 pm »
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First tip: in gravitational motion, the force on an object is not important; the object's acceleration is. As the acceleration of an object in a gravitational field is independent of its mass, any calculations of Triton's orbital radius or speed or whatever must be independent of its mass.

Second tip: don't rote learn formulas without knowing how to use them xP
Third tip: you've actually answered question a by yourself in the question :P

Here's an example of what I mean.
Let's say we want to derive the special case of Kepler's third law for circular orbits. Then, as the gravitational force provides the centripetal acceleration, we have mv^2/r = GmM/r^2
Or, if you want, v^2=GM/r
Note how the m that cancelled was the mass of the satellite and not Neptune.
Then, v=2pi*r/T, so v^2=4pi^2*r^2/T^2=GM/r
r^3/T^2 = GM/4pi^2

Do the rest yourself now
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Rishi97

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Re: Rishi's Physics Thread
« Reply #6 on: March 15, 2014, 07:35:26 pm »
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So Lzxnl, does that mean that when ever I have these sorts of questions, I ignore the mass of the actual object but instead look at the mass of the object that it is rotating around?
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lzxnl

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Re: Rishi's Physics Thread
« Reply #7 on: March 15, 2014, 09:39:38 pm »
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Yes.
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Re: Rishi's Physics Thread
« Reply #8 on: March 16, 2014, 01:07:02 am »
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I actually have no idea what I'm saying or talking about.

Rishi97

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Re: Rishi's Physics Thread
« Reply #9 on: March 16, 2014, 11:44:02 am »
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Rishi, this website helped me out as well!
http://www.physicsclassroom.com/class/circles/Lesson-4/Mathematics-of-Satellite-Motion

Thanks for that Cort. You're right, its a really good website  :D
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Rishi97

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Re: Rishi's Physics Thread
« Reply #10 on: March 16, 2014, 01:27:34 pm »
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Two identical tennis balls X and Y are hit horizontally from a point 2.0m above the ground with different initial speeds: ball X has an initial speed of 5.0m/s, while ball Y has an initial speed of 7.5m/s.
a) Calculate the time it takes for each ball to strike the ground. (I got 0.64s for both)
b) Calculate the speed of ball X just before it strikes the ground

Help with part B pls, gosh projectile motion is confusing :(
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rhinwarr

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Re: Rishi's Physics Thread
« Reply #11 on: March 16, 2014, 01:56:19 pm »
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The speed of the ball is the vector addition of the vertical velocity and horizontal velocity. Draw a triangle with the horizontal and vertical components of the speed just before the ball strikes the ground then solve for the hypotenuse (which is the speed).

Horizontal:
V=5m/s (because the horizontal velocity is constant through the flight)

Vertical:
u=0 v=? x=-2m a=-10m/s/s
V^2=u^2 + 2ax
v^2 = 2 x -10 x -2
v = 6.3 m/s

Speed = sqrt(horizontal^2 + vertical^2) = 8.1 m/s

Rishi97

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Re: Rishi's Physics Thread
« Reply #12 on: March 16, 2014, 02:05:43 pm »
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Thanks rhinwarr
Where did u get this equation from Speed = sqrt(horizontal^2 + vertical^2) ?
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Rishi97

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Re: Rishi's Physics Thread
« Reply #13 on: March 16, 2014, 02:17:06 pm »
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Can someone please help me with q8a from the 2013 physics exam? Its on projectile motion.
I'm getting the answer 1.73s but the answer says 3.0s.
HELP PLS!!!
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rhinwarr

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Re: Rishi's Physics Thread
« Reply #14 on: March 16, 2014, 06:53:13 pm »
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Where did u get this equation from Speed = sqrt(horizontal^2 + vertical^2) ?

You can use pythagoras' theorem with the vectors since they form a right angled triangle.

Quote
Can someone please help me with q8a from the 2013 physics exam? Its on projectile motion.

The information they provide allow you to calculate for the vertical component of velocity.
u = 20sin(30) m/s (vertical component of the initial velocity by resolving into a triangle)
x = -15 m (height of the cliff)
a = -10 m/s/s (due to gravity)
t = ?

Using the constant acceleration formula:
x = ut + 0.5at^2
-15 = 20sin(30)t + 0.5 x -10 x t^2
5t^2 -10t - 15 = 0
5(t^2 - 2t - 3) = 0
5 (t - 3) (t + 1 ) = 0
Time must be positive
Therefore, t = 3s