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

[huybinhs]

Homework Statement

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

Homework Equations

g = G*M/R²

a = v^2 / r

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 don't 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!

 

[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).

 

[huybinhs]

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).

"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...

 

[huybinhs]

Still don't get this. Please help!

 

[huybinhs]

Any one?

 

[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.

 

[huybinhs]

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.

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

 

[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

 

[huybinhs]

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

 

[huybinhs]

Extremely stuck with this one. Anyone?

 

[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.

 

[huybinhs]

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.

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

 

[tiny-tim]

But I couldn't find any information on the graph. It's just given 2 distances. I'm really stuck on this- my last problem. Please be more specific!

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 traveled 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.

 

[huybinhs]

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 traveled 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.

Hello,

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

 

[tiny-tim]

Hello,

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

Measure it!

 

[huybinhs]

Measure it!

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 ??

 

[tiny-tim]

(have a theta: θ )

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?

Then use v_x = 16.9 cosθ , v_y = 16.9 sinθ

 

[huybinhs]

(have a theta: θ )


Then use v_x = 16.9 cosθ , v_y = 16.9 sinθ

(have a theta: θ )


Then use v_x = 16.9 cosθ , v_y = 16.9 sinθ

Using v_x = 16.9 cosθ (i) , v_y = 16.9 sinθ (ii)

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

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

then "Then read off the horizontal distance traveled 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 = v_0x * t (1)

\Delta_y = v_0y - 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 ?)

 

[huybinhs]

Still waiting for the answer!

 

[aim1732]

Are you sure your initial angle is correct(looks like 45 to me). Delta y=0 as there is no change in height and delta x=124.

 

[huybinhs]

Are you sure your initial angle is correct(looks like 45 to me). Delta y=0 as there is no change in height and delta x=124.

We need to calculate the angle, not just guessing.

If Delta y = 0, then how can I simplify the t?

 

[aim1732]

Well I am dead sure. Look closely tan(theta)=delta x/delta y.
Square the first eqn. and eliminate t^2.

 

[huybinhs]

Well I am dead sure. Look closely tan(theta)=delta x/delta y.
Square the first eqn. and eliminate t^2.

After I square the first equation. I got:

(\Delta_x)² = (v_0x)²
0 = v_0y*(1/2) * g

then I plug in the number:

first equation = 124² = 7.57² ?

second equation <=> 15.1 * 1/2 * g = 0 now couldn't solve for the g either. I'm really stuck!

 

[aim1732]

Sorry you eqn. 2 is wrong. It is x= u*t +1/2*g*t*t. If you can take out t from first eqn. and simply put in the second eqn. you will get the result. I don't see your problem.

 

[huybinhs]

Sorry you eqn. 2 is wrong. It is x= u*t +1/2*g*t*t. If you can take out t from first eqn. and simply put in the second eqn. you will get the result. I don't see your problem.

You mean:

\Delta_x = v_0x * t (1)

\Delta_y = v_0y*t + (1/2)gt² (2)

Are those correct?

 

[aim1732]

Yes, eliminate t but don't square. Simply take t on LHS in first eqn. and put in eqn.2.

 

[huybinhs]

Yes, eliminate t but don't square. Simply take t on LHS in first eqn. and put in eqn.2.

Then I got:

t = 16.4 s

Plug t in (2), I get:

0 = 15.1*16.4 + (1/2)g*(16.4²)

=> g = -1.84 m/s² ?

 

[aim1732]

That is because your initial angle is incorrect. It is 45.

 

[huybinhs]

But how can I know the angle is 45? Please let me know!

 

[huybinhs]

That is because your initial angle is incorrect. It is 45.

If the angle is 45 then v_x = v_y = 11.95 m/s

then using those equation I got g = -2.30 m/s². Is this correct?

 

[aim1732]

Look at the graph. The gradient of the curve initially is delta x/ delta y =(20-10)/(10-0).

 

[huybinhs]

Look at the graph. The gradient of the curve initially is delta x/ delta y =(20-10)/(10-0).

I just submitted the answer -2.30 and it's INCORRECT! Please help!

 

[huybinhs]

I just submitted the answer -2.30 and it's INCORRECT! Please help!

It's all about the sign. Thanks so much for your help aim1732 and tiny-tim! I really appreciate it ;)