GREPhysics.NET: GR9677 #22

GR9677 #22

Problem

GREPhysics.NET Official Solution

This problem is still being typed.

Mechanics $\Rightarrow$ }Geometry

The harder part of this problem involves determining the radius of Mars. It's an approximate geometry problem. The problem gives a vertical drop of 2m for every 2600m tangent to the surface. The tangent to the surface is approximately one leg of a triangle whose hypotenuse is the radius of Mars, since the radius is much larger than the tangent distance. The other leg of the right triangle is just $r-2$ , where $r$ is the radius of Mars. In equation form, what was just said becomes $(r-2)^2+3600^2=r^2$ . The square terms cancel out, and dropping out the 4, one has $r\approx 3600^2/2 \approx 8E6 m$ .

(The above deduction was due to Ayanangsha Sen.)

The easier part comes in the final half of the problem: applying the centripetal force to the force of gravity. $mv^2/r = mg \Rightarrow v=\sqrt{2 g r}\approx \sqrt{20 * 8E6}=\sqrt{16E7}\approx 4000 m/s$ , which is closest to $3.6km/s$ , as in choice (C).

Alternate Solutions

Zhi Yong 2009-11-05 04:50:52	This solution may be confusing as I can't draw diagrams, but I'll try to describe it clearly. Draw a sector with an arc length of 3600 m, radius r, and angle theta. This sector has an arc and two straight lines. Create a right triangle within this sector by drawing a line from one tip of the arc to the straight line opposite. This right triangle has hypotenus of length r and one side of length r-2. Notice that cos theta = (r-2)/r = 1 - 2/r, which is approximately 1 - square(theta)/2. Therefore r=4/square(theta). Arc length s = 3600 = rtheta. Solving these two equations, r = 18001800. v = sqrt(r*g) = 3600 Reply to this comment
hamood 2007-04-09 11:49:06	setting up the problem as this really simplifies it: In time t , distance x = 3600 m, y = 2 m, a = .4g, t = 3600/v, and 2 = 1/2at^2....solve the eqs for v. Reply to this comment

Comments

Zhi Yong
2009-11-05 04:50:52

This solution may be confusing as I can't draw diagrams, but I'll try to describe it clearly.

Draw a sector with an arc length of 3600 m, radius r, and angle theta. This sector has an arc and two straight lines. Create a right triangle within this sector by drawing a line from one tip of the arc to the straight line opposite. This right triangle has hypotenus of length r and one side of length r-2.

Notice that cos theta = (r-2)/r = 1 - 2/r, which is approximately 1 - square(theta)/2. Therefore r=4/square(theta).

Arc length s = 3600 = r*theta.

Solving these two equations, r = 1800*1800.

v = sqrt(r*g) = 3600

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Albert
2009-10-22 04:42:35

With almost no calculation required, here is my answer:

For the sake of my lazyness I take g=10m/s^2 as on earth.

For mars, it is 4m/s^2, right?

Flying over Mars, I fall 4 meteres every second, if only I can fly 3600m every second, I gain 2 meters in height, thanks to the curvature of Mars.

Since I am allowed to fall 2 meters every second, I just need to fly with the speed 3600m/s or 3.6km/s and I am good.

Albert
2009-10-22 07:25:22

Make it 2 meters, where I say "I fall 4 meters".

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CaspianXI
2009-03-22 12:06:08

Yosun gets the right answer the wrong way, as ayabepaula pointed out earlier. IMO, the best solution is from adam. Since he was the first poster, his wonderful solution got buried... do yourself a favor and scroll down to the bottom of the page to see a great solution!

Yosun, I don't mean to criticize you... this is an amazing site! Thank you *so much.*

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ayabepaula
2008-10-22 10:18:58

hi !!!!\(^.^)/

Thanks for a wonderful website

1)question:
$(r-2)^{2}+(3600)^{2}=r^{2}$
when the
square terms cancel out, and dropping out the 4, one has

$r^{2} -2.2(r) +4+(3600)^{2}=r^{2}$

$4r=(3600)^{2}$

$r=3,2 \times 10^{6}$

isn't?

or

$r=8 \times 10^{6}$

i didn't understand

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Jeremy
2007-10-17 12:14:41

Wow! The kinematics approach is cool; I hadn't thought of that. Here is yet another approach, this one using similar triangles to find the radius.

Two points on the surface of Mars are connected by the two legs of a right triangle. One leg is $3600$ m long, and the other is $2$ m long. The hypotenuse is a line segment connecting the two points on the surface. The length of the hypotenuse is about $3600$ m. The similar triangle is formed with two additional line segments: one connecting the center of Mars and the midpoint of the hypotenuse, and the other connecting one end of the hypotenuse to the center of Mars. The leg resulting from the bisected hypotenuse has a length of about $1800$ m; the other leg's length is roughly the radius of Mars ( $R$ ). Proportionality of the corresponding sides gives $R/(1800 m)=(3600 m)/(2 m)=1800$ , or $R \approx 1800^{2} m$ . For the orbit, we have $g_{m}=v^{2}/R \Rightarrow v =\sqrt{g_{m}R}$ , where $g_{m} \approx 4 m/s^{2}$ . Therefore, $v \approx 3600 m/s$ (notice how easy the square root is!).

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hamood
2007-04-09 11:49:06

setting up the problem as this really simplifies it:
In time t , distance x = 3600 m, y = 2 m, a = .4g,
t = 3600/v, and 2 = 1/2*a*t^2....solve the eqs for v.

hamood
2007-04-09 11:49:48

initially the velocity is only in the x direction with magnitude v.

hamood
2007-04-09 11:50:18

initially the velocity is only in the x direction with magnitude v.

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chri5tina
2006-11-28 03:05:52

I think the above post should have been labeled as a typo alert, not NEC.

Typo: in the solution, there is an instance where it says "2600" when it should say "3600".

Additionally, in the solution, shouldn't it be that the hypoteneuse is r+2 and the two legs are 3600 and r? I made an attempt to draw a diagram from the current solution, but was unable to.

sirius
2008-11-06 22:38:53

It took me a while to figure out the picture, but here's what I've found. If you draw a circle, then on the outside surface draw what is happening. That is draw a tangent line to the circle, until you feel its length represents 3600 m. draw a line perpindicular to it back towards the circle. Draw two lines from the center of the circle: each on to where the two lines intersect the circle. Make one more line that will complete the right triangle Yosun explains.

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globalphysics
2006-11-01 21:03:27

I believe the third line should say 3600m not 2600m. Just a detail. :)

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isak
2006-10-24 11:36:42

I think there is a mistake here. 2rs=3600^2 so r=3.24E6m. Also v= $\sqrt{.4gr}$ so v=3600m/s.

ps thanks for the excellent resource

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rreyes
2005-11-27 08:49:50

In the same spirit as the above comment, for every change in height y, the ball should travel a distance R. We have $y=0.5t^2$ and $R=vt$ . Dividing, we get $v^2=0.5gR^2/y$ , where $g=4$ for Mars. After plugging in for y, we get $v^2 = R^2$ numerically so $v=3600m$ .

solar39
2009-11-06 09:06:49

Yes, one can treat this problem just like a simple trajectory problem. and solve for the speed in the x-direction with the range being 3600 meters and the displacement in the y-direction 2 meters. Just remember to change the g (gravitational acceleration) to 0.4g to get the correct answer!

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adam
2005-11-09 22:03:19

This problem easier if you notice the golf ball may fall 2m for every 3600m and still stay at the same height. The time it takes to fall 2m is:

2=1/2(0.4g)t^2

solving for t yields:

t= $\sqrt(4/.4g)$ approx=1s

now the velocity multiplied by the time it takes to fall 2m should equal 3600

vt=3600 $\Rightarrow$ v=3600/1

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Adam D
2005-11-09 21:49:30

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