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I'm new here because I think I need some help. Our teacher gave us the task to find out how fast you would have to throw something upwards (friction and mass are not considered) to get it to leave the atmosphere. So my first approach (I'm not really good at physics, don't judge me) would be just to use the formula for vertical throws that calculates height dependent on ground speed and then fill in 100km so about the distance between earth and the end of atmosphere. Leaving out mass and friction would make it really easy then, but I know that that earth's gravity force gets lower the wider something turns away from earth. So I would just use a gravity formula and a standard vertical throw formula, but is there anything else I should consider or am I thinking completely wrong?

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  • $\begingroup$ hint: can you neglect the variation of the gravitational acceleration with height? see wolframAlpha for the answer. $\endgroup$ – AccidentalFourierTransform Jan 8 '17 at 17:36
  • $\begingroup$ There are a lot of ways to approach this, each with its own set of assumptions and parameters to consider. You can help us by identifying the level of the class. Middle school? AP Physics? University intermediate mechanics ?? $\endgroup$ – garyp Jan 8 '17 at 17:36
  • $\begingroup$ High school, but ground curse, as I already said we dont consider friction or mass. So basically I would just calculate the gravitational acclearation at 50km to get the average gravitaional accleration, and use it in the formula. $\endgroup$ – Takeda43 Jan 8 '17 at 17:41
  • $\begingroup$ But what other parameters do you mean? Im really interested, as the curse has been pretty boring until this $\endgroup$ – Takeda43 Jan 8 '17 at 17:42
  • $\begingroup$ Do you know a bit of calculus? $\endgroup$ – projectilemotion Jan 8 '17 at 17:45
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If you are going to ignore friction (atmospheric drag) but not how graviational acceleration varies with distance from the Earth, use Conservation of Energy.

The potential energy $U$ of an object with mass $m$ in the central gravitational field of an object with mass $M$ is given by:

$$U(r)=-\frac{mMG}{r},$$

with $r$ the distance from the centre of the field.

By throwing up an object, kinetic energy $K$ will be converted to potential energy:

$$K=\frac12 mv^2,$$

where $v$ is the speed of the object.

Conservation of Energy tells us that:

$$\Delta K=\Delta U(r)\tag{1}$$

Say we throw the object from $r_0$ (surface of the Earth) to $r_1$, starting velocity $v_0$ and end velocity $v$, equation $(1)$ is then used to calculate $v_0$ as a function of $v$, $r_1$ and $r_0$.

Note that neglecting air drag will underestimate $v_0$ because work (kinetic energy) needs to be done against the drag force.

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  • $\begingroup$ So where is the difference between Escape velocity and this? $\endgroup$ – Takeda43 Jan 8 '17 at 18:09
  • $\begingroup$ the escape velocity is obtained by taking $100\ \mathrm{km}=\infty$. $\endgroup$ – AccidentalFourierTransform Jan 8 '17 at 18:10
  • $\begingroup$ They are very related. For escape velocity, $r_1$ is set to infinity: $r_1=+\infty$. That means that $U(r_1)=0$. $v_0$ is then the escape velocity of the object (again, ignoring air friction). Also, set $v=0$ (zero end velocity). $\endgroup$ – Gert Jan 8 '17 at 18:11

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