Physics problem involving a gravitational change based on distance traveled.

[ejezisek]

Homework Statement

An object is launched at a speed of 10.1km/s. What is the maximum height of this object.
The answer is within 10% of 2.88x10^7. and should be relatively close to it.

Homework Equations

radius of earth=re.
I believe 9.8*((re)/re+h)^2=gravity is relevant. a=g. v=gx-10100. and d= gx^2/2-10100x. There may be others but I do not know them.

The Attempt at a Solution

I tried factoring out re+h^2 and plugging in values but I was unable to figure out if there were any other formulae necessary. The problem also gave practice problems and i tried using a table to find an approximate value for h. I tried determining a formula from these answers but was unable to.

 

[ehild]

You can not assume constant gravitational force.

ehild

 

[ejezisek]

I thought I might be able to find the average gravitational force used in the equation. And use that for g. Is this wrong? The main thing I am looking for is additional formulae I may be missing.

 

[zachzach]

Can you use energy? You cannot find the average acceleration without knowing the height.

 

[ejezisek]

I am not entirely sure on how to do that. Could you please explain?

 

[zachzach]

I did it with energy but am within 20% of the answer so I am not sure now.
$$E_i = K_i + U_g$$
$$E_f = U_f$$
What is the equation for gravitational potential energy?

 

[ejezisek]

Me=Mass of the earth
m=mass of object away from the Earth which we don't know.
r= the distance halfway between the middle of the Earth and the middle of the object.
G= gravitational constant 6.673* 10^-11.
U(r)=-G*Me*m/r
Thank you very much for trying/helping me out :)
If it would help you out i could provide a sample problem with the answer.

 

[ehild]

Using constant g is wrong for a distance comparable or greater than the radius or Earth. The launch speed is very close to the escape velocity, so you can expect that the object flies much further than the Earth radius.
In the formula for the gravitational energy, r is the distance of the object from the centre of Earth. Not halfway.
Use the gravitational potential energy at the surface of Earth + the initial KE to obtain the total energy of the object, and then find the distance from the centre of Earth where the potential energy is equal to the total energy, that is, the kinetic energy is zero.

ehild

 

[zachzach]

Me=Mass of the earth
m=mass of object away from the Earth which we don't know.
r= the distance halfway between the middle of the Earth and the middle of the object.
G= gravitational constant 6.673* 10^-11.
U(r)=-G*Me*m/r

r = distance between center of mass of Earth (~center of sphere) the object. So on surface of the Earth $$U_g = \frac{-GMm} {R_E}$$.

Do you think that energy is conserved, because I don't see any reason why it should not be. so write out the initial and final energies.

 

[ejezisek]

At the Earth's surface a projectile is launched straight up at a speed of 10.0 km/s. To what height will it rise? Ignore air resistance and the rotation of the Earth.
2.52 * 10^7 is the answer.

 

[ejezisek]

How is it that i calculate the final energy?

 

[zachzach]

How is it that i calculate the final energy?

What is the velocity? What is the potential energy?

 

[ejezisek]

how do i calculate kinetic energy without knowing the mass of the object?

 

[zachzach]

Write out your entire equation on here.

 

[ejezisek]

1/2 m * v * v I'm trying to do what ehild said above use potential and kinetic energy. I guess i probably don't need to know m. Ill just try using m as 2 for both potential and kinetic.

 

[zachzach]

If you do this problem right you won't need m. You cannot just arbitrarily say m= 2 btw. You wrote down the kinetic energy only.

1.What is the INITIAL total energy (kinetic + potential)?

2.What is the Final energy once the object reaches it's max height.

3.E(1) = E(2) :conservation of energy.

 

[ejezisek]

i can pretend the object has any mass i want. the initial energy is 102010000 for an object with m=2kg. The initial potential energy would be 0 I believe.

 

[zachzach]

lol. I have already told you what the initial potential energy is in a previous post. What is also funny is if you are going to pretend you know the mass why not choose m =1 lol.

 

[ejezisek]

i chose m=2 because i have to divide by two in the kinetic energy equation.
kinetic energy:102010000
total energy: -22998502
PE:-125008502

 

[zachzach]

I don't know if you are right. If you write the EQUATION out it is much easier. Don't feel like doing math with large numbers since I am terrible at it anyways.

 

[ejezisek]

I'm sure I messed up somehow i got 1.73*10^7m and well... that can't be the answer.

 

[ehild]

Did you calculate with that 2 kg again?

ehild

 

[ejezisek]

m=2
Kinetic Energy=1/2 m * v^2=v^2=10100^2=(1.01*10^4)^2=1.02*10^8
Potential Energy=-GMem/r=-6.673*10^-11*5.97*10^24*2/6.38*10^6=-10^7*6.673*5.97*2/6.38=-12.5*10^7=-1.25*10^8
Total Energy=-2.3*10^7
Potential Energy*earthRadius/-2.3*10^7=radius
-1.25*10^8*6.38*10^6/-2.3*10^7=1.25*6.38*10^7=7.975*10^7
7.975*10^7-.638*10^7
7.337*10^7 Not the right answer will do it again...o didnt divide by -2.3.

 

[ejezisek]

I GOT IT! Thanks for helping :)

 

[zachzach]

woo hoo! I hate working with big numbers.

 

[ehild]

-1.25*10^8*6.38*10^6/-2.3*10^7=1.25*6.38*/-2.3*10^7=3.467*10^7

You left the total energy somewhere:). Subtract the radius of Earth, as the height above the surface was the question.

ehild