Evaluate Magnitude of Gravitation at the Surface of a Planet

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Homework Help Overview

The discussion revolves around evaluating the magnitude of the gravitational field at the surface of a planet based on the trajectory of a rock thrown from a height with an initial speed of 20.9 m/s. The problem is situated in the context of projectile motion and gravitational effects, with the planet assumed to have no atmosphere.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the trajectory of the rock and the gravitational field, questioning how to interpret the graph and the meaning of various components such as the launch angle and maximum height. There are attempts to apply trigonometric relationships and projectile motion equations, with some participants expressing confusion about the necessary measurements and calculations.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and the equations involved. Some guidance has been offered regarding the use of projectile motion equations and the importance of accurately reading values from the graph. However, there is no explicit consensus on the best approach or the specific values to use for calculations.

Contextual Notes

Participants note challenges in obtaining precise measurements from the graph, which may affect their calculations. There is also mention of homework constraints that may limit the methods available for solving the problem.

ScrubTier
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1. The trajectory of a rock thrown from a height with an initial speed of 20.9 m/s is shown in the figure below. Evaluate the magnitude of the gravitational field at the surface of the planet. The planet has no atmosphere.
https://s2.lite.msu.edu/cgi-bin/plot.png?file=muiblack_msu_1442688985_1543723_plot.data

Homework Equations



Tangent = Opposite/Adjacent
Xmax=Vo^2/a*sin2Theta=Vo^2/a*2sinTheta*cosTheta[/B]

The Attempt at a Solution



Rise = 15 Run=20
Tan(x)=15/20
x=36.869 I think. I got this number but plugging this back in is not working for me. I thought I would take this and plug it into a triangle where h=20.9 (Vo) then the y-component would be the answer. This is not working.[/B]
 
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ScrubTier said:
1. The trajectory of a rock thrown from a height with an initial speed of 20.9 m/s is shown in the figure below. Evaluate the magnitude of the gravitational field at the surface of the planet. The planet has no atmosphere.
https://s2.lite.msu.edu/cgi-bin/plot.png?file=muiblack_msu_1442688985_1543723_plot.data

Homework Equations



Tangent = Opposite/Adjacent
Xmax=Vo^2/a*sin2Theta=Vo^2/a*2sinTheta*cosTheta[/B]

The Attempt at a Solution



Rise = 15 Run=20
Tan(x)=15/20
x=36.869 I think. I got this number but plugging this back in is not working for me. I thought I would take this and plug it into a triangle where h=20.9 (Vo) then the y-component would be the answer. This is not working.[/B]
What is x supposed to be? What units does it have? What does "evaluate the magnitude of the gravitational field" mean to you?
 
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x is supposed to be the angle at which the rock is first thrown. It would be degrees. The magnitude of the gravitational field would be the constant y-component wherever the rock is on the graph.
 
ScrubTier said:
x is supposed to be the angle at which the rock is first thrown. It would be degrees. The magnitude of the gravitational field would be the constant y-component wherever the rock is on the graph.
No, I don't think so. According to the graph of the trajectory, there is no constant y-component. It keeps changing ...

Since this is a planet which is presumably not earth, I think the "magnitude of the gravitational field" means finding the value of g for this planet.
 
But how would I do that then? I'm so lost.
 
ScrubTier said:
But how would I do that then? I'm so lost.
Write out the equations for projectile motion under a constant gravitational acceleration. You'll use the information from the graph of the trajectory to help you solve for the value of g for this planet.
 
Is it like the equation I included?
 
ScrubTier said:
Is it like the equation I included?
No.
The usual SUVAT equations allow you to write out the horizontal and vertical displacements as functions of time. But your graph says nothing about time. It shows a trajectory, i.e. the relationship between x and y. If you have an equation showing how x depends on t, and another showing how y depends on t, how would you obtain an equation that shows how y depends on x?

That said, there is a way to use the equation you posted. First, you have to understand what exactly xmax refers to in that equation and read that off the graph. You also need to ead theta from the graph, but it will be hard to do that with any accuracy. So I recommend working with the SUVAT equations as I described above.
 
Would I need Xmax and Ymax to use those equations? My professor said we should be able to read those from the graph but it isn't so clear..
 
  • #10
What I have figured out is that I can find the x-component of Velocity by my given velocity (20.9) times cos(Launch angle) and my y-component by 20.9*sin(launch angle) but how do I find the launch angle? Can I use an angle found on the downslope?
 
  • #11
ScrubTier said:
What I have figured out is that I can find the x-component of Velocity by my given velocity (20.9) times cos(Launch angle) and my y-component by 20.9*sin(launch angle) but how do I find the launch angle? Can I use an angle found on the downslope?
It's not clear what you mean by "downslope" here.

Since you are given a plot of its trajectory, you can find (or approximate) the angle which the projectile takes when it is launched.
 
  • #12
I don't understand how to do that.
 
  • #13
ScrubTier said:
I don't understand how to do that.
Well, take the sketch and try to draw a tangent to the curve at the start of the trajectory. The angle this tangent line makes to the horizontal will be launch angle, or a close approximation.
 
  • #14
I know how to find the angle I just don't know what measurements to use. Would x=0, 20 y=15, 30 be close enough?
 
  • #15
I just tried by using those numbers. It was wrong. I know I must use something to find the Launch Angle then find x-component of velocity and y-component of velocity. Then determine the Xmax (distance which it traveled when it made it back to the starting height (15m)) and Ymax (max height it got to). I use these in d(Xmax)=Vo(x)*t to find t then d (which is 0) = Vo(y)*t-1/2*g*t^2
I just do not know what numbers to use to find these things!
 
  • #16
ScrubTier said:
I just tried by using those numbers. It was wrong. I know I must use something to find the Launch Angle then find x-component of velocity and y-component of velocity. Then determine the Xmax (distance which it traveled when it made it back to the starting height (15m)) and Ymax (max height it got to). I use these in d(Xmax)=Vo(x)*t to find t then d (which is 0) = Vo(y)*t-1/2*g*t^2
I just do not know what numbers to use to find these things!
Well, we can't help you if you don't post your calculations. You may be making mistakes with your artihmetic
 
  • #17
If you look at the picture of the graph finding coordinates to do calculations with is very un-exact. That is my problem.
 
  • #18
ScrubTier said:
If you look at the picture of the graph finding coordinates to do calculations with is very un-exact. That is my problem.
Yes, I understand that.

But what are we to suggest to you in response to the statement, "I got it wrong"?
 
  • #19
I really want advice as to what numbers I should use for my equations. Estimates to what Xmax should be would be helpful.
 
  • #20
ScrubTier said:
I know how to find the angle I just don't know what measurements to use. Would x=0, 20 y=15, 30 be close enough?
That looks somewhat off to me. As I posted, I think that estimating the launch angle directly from the graph is a rather inaccurate way to proceed. You would be much better off picking a point on the trajectory that is clear (max height gain and x at that point) and finding the gravitational acceleration and angle which hit it.
 
  • #21
Reading points from a graph is easier and usually more accurate than trying to estimate a slope by eye, particularly when the axes have different scales. I've taken your figure, added some additional rulings to the axes to aid in estimating readings, and indicated a few points that you might find useful in working the problem:
Fig3.gif
 

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