Evaluate the magnitude of the gravitational field at the surface of the planet

1. Feb 26, 2010

huybinhs

1. The problem statement, all variables and given/known data

The trajectory of a rock thrown from a height with an initial speed of 16.9 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.

The graph is as follow:

http://i995.photobucket.com/albums/af79/huybinhs/plot-1.png

2. Relevant equations

g = G*M/R²

a = v^2 / r

3. The attempt at a solution

I know using g = G*M/R² to find magnitude of the gravitational field. In order to find R using a = v^2 / r; but I dont know how to find a from the graph. Please help! Thanks!

Note: I'm not sure if my tried solution is correct or not. Please advise!

2. Feb 26, 2010

tiny-tim

Hi huybinhs!

Sorry, but you're going completely berserk on this.

You don't know the mass of the planet, or its radius, so you can't use the first equation; and the second equation is for centripetal acceleration

All you're being asked is to measure the acceleration on the graph (you can assume it's vertical, and a constant).

3. Feb 26, 2010

huybinhs

"All you're being asked is to measure the acceleration on the graph (you can assume it's vertical, and a constant)" => you mean a = 10 m/s^2 (when I look at the vertical on the graph) ??? I'm confused...

4. Feb 26, 2010

huybinhs

5. Feb 26, 2010

huybinhs

Any one?

6. Feb 27, 2010

mgookin

I haven't looked at your graph, but acceleration in any equation is the second derivative. First derivative is velocity, second is acceleration.

So if your graph shows you something happening at the surface of the planet, take the second derivative of the function at that place on the chart.

7. Feb 27, 2010

huybinhs

I understand. My graph showing Vertical Positive Above Surface (m) and Horizontal Position (m). What can I do?

8. Feb 27, 2010

Mindscrape

Tell me, between which of vertical position and horizontal position does gravity affect? Also this equation may help you

v^2=v0^2+2as

9. Feb 27, 2010

huybinhs

The gravity absolutely affects on the rock. Using your formula, what about the v?

Last edited: Feb 27, 2010
10. Feb 28, 2010

huybinhs

Extremely stuck with this one. Anyone????

11. Feb 28, 2010

willem2

Suppose you throw away a rock from an alitude of 10m with a speed v=16.9m and with an angle of $\phi$ with the horizontal on a planet with acceleration of gravity a.
you should be able to calculate the maximum altitude and the range as a function of a and $\phi$

Now you can read the maximum altitude and the range from the graph, so you get to equations
for a and $\phi$ that you can solve.

12. Feb 28, 2010

huybinhs

Got ya. Do you have any ideas about equation calculator here? That's would be great ;)

13. Feb 28, 2010

tiny-tim

Hi huybinhs!

From the graph, measure the initial angle, and use that to find the initial components of velocity (you know the magnitude is 16.9).

Then read off the horizontal distance travelled when the ball returns to its original height (10m).

Call the time when that happens "t".

The use the standard constant acceleration equations (with a = 0 and a = -g) for the x and y directions …

that gives you two equations, from which you can eliminate t and find g.

14. Feb 28, 2010

huybinhs

Hello,

How can I find intial angle on the graph???? guessing?

15. Feb 28, 2010

tiny-tim

Measure it!

16. Feb 28, 2010

huybinhs

Ok. Taking x = 0; x =10 ; y = 10 ; y = 20. So:

tan(theta) = o / a = 20 /10 = 2 => angle = 63.4 degree. Correct? then what?

what formula to calculate the v?

I got v^2 = tan(theta) * r*g but lack of r and need to find g ???????

Last edited: Feb 28, 2010
17. Feb 28, 2010

tiny-tim

(have a theta: θ )
Then use vx = 16.9 cosθ , vy = 16.9 sinθ

18. Feb 28, 2010

huybinhs

Using vx = 16.9 cosθ (i) , vy = 16.9 sinθ (ii)

(i) => vx = 16.9 * cos (63.4) = 7.57 m/s

(ii) => vy = 16.9 * sin (63.4) = 15.1 m/s

then "Then read off the horizontal distance travelled when the ball returns to its original height (10m)." = 124 m

"Call the time when that happens "t"."

Using the standard constant acceleration equations (with a = 0 and a = -g) for the x and y directions:

$$\Delta$$x = v0x * t (1)

$$\Delta$$y = v0y - 1/2 gt2 (2)

How can you cancel the t because the equation (2) having t2 ; what's value for $$\Delta$$x and $$\Delta$$y ?? (60 and 124 ????)

19. Mar 1, 2010