What method should I use to solve for the final acceleration of a projectile being launched from the earth's surface?

The question I am working on is:

A projectile is launched vertically from the surface of the earth with an initial speed $v$. What is the magnitude of its acceleration $a$ ($\mathrm{m}/\mathrm{s}^2$) $d$ km above the surface of the earth?

I cannot use a formula for an object under constant acceleration because the force between the object and the earth changes at large distances. My textbook has formulas for the universal law of gravitation and also for gravitational potential energy. I am thinking that I may need to solve this problem using the energy formulas. The formula given for potential energy is: $$U(r) = - {GM_Em}/{r}$$

Using this I can solve for the final potential energy. How do I derive the final acceleration from this?

  • 1
    $\begingroup$ Well, for starters, the (magnitude of the) acceleration would crucially depend on how far above the surface of the Earth that the projectile is. $\endgroup$
    – Qmechanic
    Jul 25 '11 at 22:23
  • $\begingroup$ You were right. I neglected to specify that the question does include $d$, the distance from the surface of the earth. $\endgroup$ Jul 28 '11 at 4:21

I did the problem, but it would appear I answered a little too much, so here is a bit of a scaled down version. Firstly, I'm going to try to echo back the question to make sure we're talking about the same thing.

A projectile is launched vertically from the surface of Earth at a speed v. Find the velocity at the highest point on its trajectory without assuming constant gravity.

The reason I had to add that last part is because it was never really specified where along the path the acceleration is needed.

In short, yes, an energy balance is needed in order to identify the properties of the highest point it reaches, in particular, $r$, which is the distance from the center of the Earth. Similar to all free-fall problems, the mass of the projectile $m$ doesn't matter, so it is a good guess that eliminating that from the energy balance will be fruitful.

$$\frac{1}{m} \left( KE = PE_{final} - PE_{initial} \right)$$

It is probably obvious that the form of potential used in the question will have to be employed. It will also help to note that we are generally assuming $M\gg m$, where $M$ is the mass of Earth itself. Then just know the form of Newtonian gravity itself, $F=\frac{GMm}{r^2}$, and you're pretty much there.

  • $\begingroup$ Ok. What I am confused about now is how is $F = {GMm}/{r^2}$ going to help me if the unknowns in your equation are $m$ and $v_f$. I can solve for $v_f$, but $m$ is what I need to plug into $F = {GMm}/{r^2}$, right? So how do I solve for $m$ when it has been factored out of the equation? $\endgroup$ Jul 28 '11 at 5:25
  • $\begingroup$ In the problem I posed $v_f=0$. Now, if this is not what you're looking for, then you need to pose a clear problem, but the projectile is fired straight up and the tip top of it's path has zero velocity. You should eliminate $m$ in all relevant equations. $F$ is only written so that you can get $a$, which is what is asked for. $\endgroup$ Jul 28 '11 at 13:54
  • $\begingroup$ For a more general solution, try physics.stackexchange.com/questions/227391 $\endgroup$
    – user854
    Jan 4 '16 at 16:03

It looks like you've already found the final distace through the use of the kinetic energy forumla and the gravitational potential energy formula.

To find the acceleration, you need to use newton's second law: $$F=ma$$ and Newton's law of gravity: $$F=\frac{Gm_1m_2}{r^2}$$

Plug one of the equations into the other, solve for $a$, and you have the answer.


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