# 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

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!

Homework Helper
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

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?

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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 ;)

Homework Helper
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?

Homework Helper
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 ??

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Homework Helper
(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 vx = 16.9 cosθ , vy = 16.9 sinθ

huybinhs
(have a theta: θ )

Then use vx = 16.9 cosθ , vy = 16.9 sinθ

(have a theta: θ )

Then use vx = 16.9 cosθ , vy = 16.9 sinθ

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 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 = 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 ?)

huybinhs

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 = (v0x)2
0 = v0y*(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 = v0x * t (1)

$$\Delta$$y = v0y*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 vx = vy = 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).