# Physics Question - Centripetal Force

Discussion in 'MCAT: Medical College Admissions Test' started by y0ter, Feb 15, 2007.

1. ### y0ter New Member 5+ Year Member

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Hey folks, I have a question regarding centripetal acceleration (v^2/r) and would appreciate it if someone can take a look at the following and let me know if my understanding needs to be corrected. particularly with velocity.
I understand that centripetal force is the force necessary to allow an object to move in a circle with radius r.
In the case of a sattelite orbiting the earth v^2/r must equal the gravitational acceleration in order to avoid falling to earth correct? which means that there is ONLY one particular velocity that will allow the satellite to remain in orbit (v^2/r must equal GMe/r^2). any slower will cause the satellite to fall to earth. any faster and the satellite will exit the orbit, tangential to the orbit.

In the case of a car rounding a curve since the frictional force exerted on the tires is dependent on the velocity of the car, as long at the car's velocity does not exceed the maxiumum frictional force (static friction x Normal Force) the car will stay on the road.

Also, I encountered the following two questions and find the reasoning behind the two answers puzzling.

Q1.
Velocity of satellite is 28000 km/hr at 1600 km above the earth's surface. What is the velocity of at fighter jet 1000km above the earth's surface?
The answer choices given were 17,500 km/hr, 28,000 km/hr, 44,500 km/hr and 56,000 km/hr. The answer choice was 28,000, the reason being that the difference in gravitational acceleration at 1600 km and 1000 km is negligible. Given the answer choices I would have to agree that 28,000 is correct. In reality the fighter jet would have to have a greater velocity than 28000 km/hour correct? Or if one were to assume that gravitational acceleration is negligible velocity would be less for the fighter jet
[(28000km)^2/1600 km] = v^2/1000 km.

Q2.
2 lead sphere's, one placed 1 meter above the other sphere, is dropped at the same time. What will happen to the distance between them as they fall.
a) Remain Constant
b) Distance between the spheres will increase
c) Distance between the speres will decrease
d) cannot recall.

Answer was choice b. I can understand the reasoning behind the answer F=Gm1Me/r^2. But per question 1 if the difference in gravitational acceleration is negligible for two objects at 1600 and 1000 km, wouldn't the difference of g for 2 objects spaced 1 meter apart be even more negligible?

Thanks!

2. ### DrBowtie Final Countdown Moderator Emeritus 10+ Year Member

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#2 has nothing to do with F=Gmm/r^2. Its a d= vt+1/2gt^2 question.

3. ### killinsound Physician 10+ Year Member

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actually, it would have everything to do with F = GMM/r^2 if they say the answer is B. if they are saying its negligible, then you are right. However, under your assumption the answer would be A.

4. ### AlternateSome1 Burnt Out Physician 10+ Year Member

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Q1: Two problems, only one of which is significant to your problem. First, you cannot set the two halves of the two equations equal to each other without first getting the other half of the equations equal to each other (I'm sure that was clear.) As a demonstration...

Equation 1: (28000^2) / r1 = G * M / (r1^2)

Equation 2: (v ^ 2 ) / r2 = G * M / (r2^2)

If the two radii are not equal, then G * M / (r1^2) is not equal to G * M / (r2^2) so you cannot set the other half of the equations equal to each other. However, solving for the equations only leads to the two halves being multiplied by the radius rather than divided by it, like so: (28000^2) * r1 = (v^2) * r2.

Even though this error may be significant in other problems, it does not alter the over all concept (and thus the real point of the question). The bigger issue is the fact that you do not have the correct radius in either equation. The radii are not 1600 and 1000, they are the radius of the earth PLUS the distance above the earth. This results in radii of 6378 + 1600, or 7978, and 6378 + 1000, or 7378. Now, solving for the equation:
1) 28000^2 * 7978 = v ^2 * 7378
2) 847757115.7 = v^2
3) v = 29116 (approximately)

This makes the distance from the earth less important and makes the question a little more reasonable.

Q2: This seems a little tricky. The distance between the balls increases. Yes, it is negligible, however, A is incorrect. So...B is correct, the distance between them will increase due to the ball closer to the earth always experiencing slightly higher gravitational pull, but it is a very tiny amount.

5. ### DrBowtie Final Countdown Moderator Emeritus 10+ Year Member

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Unless they are extremely massive spheres, the force between them should be negligible. Truthfully, I didn't look at the question too closely.

6. ### y0ter New Member 5+ Year Member

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alternate, thanks for your explanation. i didn't solve the equations properly. as for the initial statements are they correct or do they require any fine tuning?

thanks.

7. ### AlternateSome1 Burnt Out Physician 10+ Year Member

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Yes, it sounds like you have a good understanding of the concepts. Now march on to rock the PS section. 8. ### y0ter New Member 5+ Year Member

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thanks alternate!

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