I don't know how to start, but i will give you some context: Let's say that we have a body that we want to vertically propel at a distance of $h_f=10$ km of it's original spot considering that the altitude from which we trow it is $h_0=0$ and the radius of the earth is $R=6371$ km. So, we can obtain the initial gravitational acceleration experimented by the body of $g_0=9.81365 \frac{m}{s^2}$ . If we consider this initial gravitational acceleration is constant while the distance from the body to the center of gravity rises, then we can further calculate the initial velocity at which we have to throw the object to rise 10 km using the the law of conservation of energy obtaining: $$\frac{mv_0^2}{2}=m\cdot g_0 \cdot h_f \implies v_0\simeq{443}\frac{m}{s}$$

Now, the question comes: Let's say that we throw the same object with the same initial velocity, but consider that at the smallest time interval($\Delta t\to 0$) or at the most distance interval($\Delta d \to 0$), momentarily will stuck with the last, the gravitational acceleration changes because the body rises from its original spot and furthers from the earth, then what is the actual height that it reaches?

I made a program in which you establish the intended height to be reached with a constant gravitational acceleration from which to be calculated the initial speed of propulsion and a distance interval at which to take into account modifications in the gravitational acceleration to obtain the actual height reached with the same initial speed. Finally for the example I gave, I observed the smaller the distance at which the change in gravitational acceleration is taken into account, more specifically 0.1 m, the current height will tend to a value in our case to $10.015...$ km and something, $\approx 15$ m. I know that is insignificant for small speeds, but is there an actual formula for this case? I will let you some equations that i work with and if you want the code(c++) just ask me.

$$g \left( h_k \right) = g_0 \cdot \frac{\left( R+h_0 \right)^2}{{\left( R+h_k \right)^2}}= g_0 \cdot \frac{R^2}{\left( R+h_k \right)^2}$$ $$d_k = R + h_k$$ $$g_{average} = \frac{|v_f-v_0|}{t} = \frac{v_0}{t} \text{,} v_f = 0 \frac{m}{s} \text{(1)}$$ $$\left. \begin{array}{l} v_{average} = \frac{h_f - h_0}{t}\\ v_{average} = v_0 - g_{average} \cdot \frac{t}{2} \end{array} \right\} \implies \frac{v_0}{2} = \frac{h_f - h_0}{t} \implies h_f = \frac{v_0 \cdot t}{2} + h_0$$ $$A_{average} = \frac{g_f -g_0}{t} = \frac{g_0 \cdot \frac{d_0^2}{d_f^2} - g_0}{t} = \frac{g_0 \cdot \left( d_0^2-d_f^2 \right)}{d_f^2 \cdot t} \text{(average supraacceleration of gravitational acceleration)}$$ $$g_{average} = g_0 + a_{average} \cdot \frac{t}{2} = \frac{g_0 \cdot \left( d_0^2+d_f^2 \right)}{2 \cdot d_f^2} \text{(2)}$$ $$\text{Consider } h_k = h_0 + k \cdot \Delta d \implies d_k = d_0 + k \cdot \Delta d \text{ and } g_k = g_0 \cdot \frac{d_0^2}{d_k^2} \text{, where } n \cdot \Delta d = h_f \text{ .}$$ $$\frac{m \cdot v_0^2}{2} = m \cdot g_0 \cdot \Delta d + m \cdot g_1 \cdot \Delta d +\text{...}+ m \cdot g_{n-1} \cdot \Delta d \implies \frac{m \cdot v_0^2}{2} = m \cdot g_0 \cdot \Delta d \cdot d_0^2 \cdot \sum_{i=0}^{n-1} \frac{1}{\left( d_0 + i \cdot \Delta d \right)^2}$$

The problem is that I basically work with two unknowns, space and time and what i reach if I equal the first $g_{average} \text{(1)}$ with the second $g_{average} \text{(2)}$ and replace $h_f \text{ with }\frac{v_0 \cdot t}{2} + h_0$ and further make calculus i reach a cubic equation with the single unknown, t.

Only, hope that it's not a stupid question! Also, what happens if instead of taking into account the smallest space interval($\Delta d \to 0$) we take the smallest interval of time($\Delta t \to 0$)?

  • 1
    $\begingroup$ Why not just use the law of conservation of energy again? We know that $U_\text{grav}=-GMm/R$. You might also be interested in escape velocity $\endgroup$ Commented Dec 13, 2022 at 17:32
  • $\begingroup$ It seems that you are asking a question about numerical integration. I would recommend that you try to formulate your question in the simplest possible terms, with that in mind. For example, does your suffix 'i' refer to subsequent time points? Since g is a function of height, $g_i$ seems like a strange variable to introduce. If you do the numerical integration properly, even a relatively large step should give you the "right" answer; if your answer depends strongly on step size, you probably are using an inappropriate numerical approximation. But to figure that out, simplify first. $\endgroup$
    – Floris
    Commented Dec 13, 2022 at 20:06
  • $\begingroup$ Yeah I'm sorry what i meant from $g_i$ is $g_0$ or $g_{initial}$. I edited the question and also expanded the law of conservation of energy in the second example to be more understood. What I mean is if we consider modification in gravitational acceleration at every $\Delta d \to 0$ than $h_f \to$ $10.015$ km, instead of $10$ km intended if consider gravitational acceleration constant. $\endgroup$
    – user353361
    Commented Dec 14, 2022 at 6:13
  • $\begingroup$ All the varying 'g' values is actually encoded in a = GM/r^2. If you input the initial conditions properly, radius of the earth & specific starting location of the object in [x,y,z] coordinate, then find the new position of the object every second from 0:00, 0:01, 0:02, 0:03... and record the maximum r attained through the whole trajectory, you will have the maximum height, plus the location of its fall back to earth, and much more. There is no need to calculate anything else such as momentum, energy. while doing the simulation $\endgroup$
    – James
    Commented Dec 14, 2022 at 7:48
  • $\begingroup$ $g\left( h\right) =\dfrac{g_{0}}{\left( 1+\dfrac{h}{R}\right) ^{2}}$ $\endgroup$
    – Eli
    Commented Dec 14, 2022 at 13:53

1 Answer 1


Lets start with the Newton gravitation law at the earth surface

$$m\,g_0=\frac{m\,M\,G}{R^2}\quad\Rightarrow\quad g_0=\frac{M\,G}{R^2}$$

and at h meter above the earth surface

$$g(h)=\frac{M\,G}{(R+h)^2}=\frac{M\,G}{\left[R(1+\frac hR)\right]^2}= \frac{g_0}{(1+\frac hR)^2}$$

the equation of motion

$$\ddot h=-g\,(h)$$

with the initial condition $~h(0)=0~,\dot h(0)=v_0~$ you can solve (numerically) the above differential equation and obtain the solution h(t)

you can solve this problem also with the energy conservation

E=total energy= kinetic energy + potential energy = constant

$$ E=\frac 12 \dot h^2+U(h)\\ U(h)=-\int\,g(h)\,dh=-\frac {g_0\,R}{1+\frac hR}\quad\Rightarrow\\ E=\frac 12 \dot h^2-\frac {g_0\,R}{1+\frac hR}$$

at $~t=0,~E_0=E(h=0~,\dot h=v_0)$ $$E_0=\frac 12\,v_0^2+g_0\,R=\rm constant$$

the maximal height is achieved when the velocity $~\dot h~=0~$

thus , solving the equation $~E(\dot h=0)=E_0~$ for h you obtain the maximal height.

$$h=\frac{v_0^2\,R}{2\,g_0\,R-v_0^2}\quad,v_0^2 > 2\,g_0\,R$$

this is the exact solution

  • $\begingroup$ Yeah, I think that I figured it out somewhat. I wrote $$g_f = \frac{g_0 \cdot \left(3 \cdot h_f^2 - 2 \cdot R \cdot h_f + R^2 \right)}{R^2} \text{. Used binomial expansion!}$$ $$g_{average} = \frac{g_0 \cdot \left(3 \cdot h_f^2 - 2 \cdot R \cdot h_f +2 \cdot R^2\right)}{2 \cdot R^2}$$ $$\frac{m \cdot v_0^2}{2} = m \cdot g_{average} \cdot h_f \implies \frac{v_0^2 \cdot R^2}{g_0} = 3 \cdot h_f^3 - 2 \cdot R \cdot h_f^2 +2 \cdot R^2 \cdot h_f$$. Above we have a cubic equation(I know that it's strange), where $h_f$ has 2 imaginary solutions and 1 real, the real one is $h_f \approx 10015.48178 m$ $\endgroup$
    – user353361
    Commented Dec 15, 2022 at 15:45
  • $\begingroup$ Which is $15m$ more from what will you obtain if you consider that all along the way, while it rises the initial gravitational acceleration remains constant. I also know that the difference in the results is insignificant for the huge velocity that I took into account and that for small velocities which are usually worked with, the difference is more insignificant. Does it have a logic because I'm in early highschool and I really want to understand physics? Nevertheless, thanks for the answer. $\endgroup$
    – user353361
    Commented Dec 15, 2022 at 15:46
  • $\begingroup$ I see, which data you used $~v_0~,R~,g_0~$ ? $\endgroup$
    – Eli
    Commented Dec 15, 2022 at 16:33
  • 1
    $\begingroup$ Oh, I understand know! Thank you very much for your answer, really helpful! $\endgroup$
    – user353361
    Commented Dec 15, 2022 at 19:47

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