For the past week, I've been working on an Effective Gravitational Acceleration calculator on Desmos. This calculator ignores drag and assumes that a rocket is flying up completely straight.

What this calculator does, is that it takes an input (of altitude above the Earth's surface), and then calculates what the acceleration of gravity (in seconds squared) would be on the interval from said altitude above Earth to the ground. This is done by taking the integral of the area under the cure that is the acceleration of gravity between the surface and the inputted altitude, and then divides it by the change in altitude.

After this, the effective gravitational acceleration is inputted into some freefall equations, and we can then calculate what speed an object would reach if it was dropped from the inputted altitude to the ground (ignoring the atmosphere and terminal velocity etc.).

In theory, this resulting total velocity of gravity is what a rocket flying straight up would need to achieve to reach the inputted altitude.

However, when I input an altitude of 160 km, I get an output of around -1700 m/s of gravitational loss to reach that altitude, while online sources say that it should be around or less than -1500 m/s.

Where am I making a mistake? Is there something wrong with my equation setup? Or, am I using a flawed logic.

Here is the calculator: https://www.desmos.com/calculator/9k3etkwmol

Here are the equations I used:

altitude (a) = 160,000

radius (r) = 6,371,000 + a

Gravity of x [G(x)] = ((6.6725910^-11)(5.9721910^24))/x^2

Delta x [dt(x)] = r - (r - a)

Average G(x) (C) = (int. [r-a, r] G(x) dx)/dt(x)

C is the average velocity of gravity.

time (t) = -(sqrrt.(2a/-C))

positive total velocity (w) = Ct

total velocity (Tv) = -w

Tv is the total velocity that an object would achieve if dropped from 'a' altitude.

By the way, I don't know how to format math like this. I tried my best.

Thank you!

  • $\begingroup$ Edited to put the correct calculator link in. $\endgroup$
    – user48194
    Aug 28, 2022 at 0:34
  • $\begingroup$ Some leading questions: What does your calculator output if you hold the acceleration constant at 9.81 m/s^2? How much different is the answer if you allow for the variation of gravity with altitude? What results do you get if you run a simple rectangular integrator vs time for the same two scenarios? And what are the "online sources" you mention? Hint: at an altitude of 160 km the acceleration of gravity has only dropped about 5% from the sea level value. $\endgroup$ Aug 28, 2022 at 1:50
  • $\begingroup$ This calculator simply calculates how much an object would accelerate if it were dropped from a certain height above earth (accounting for the drop of gravity with altitude). The answer quite honestly is not much different with the variation of gravity, because as you mentioned, gravity has dropped only marginally (around 0.5 m/s^2 i believe), By "online sources", I really meant that most websites cite that rockets tend to have a gravitational loss of about -1500 m/s, while my method is calculating -1700 m/s. $\endgroup$
    – user48194
    Aug 28, 2022 at 2:50
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    $\begingroup$ Nice job! In fact you can just use the simple 'motion at constant accel' equations and get really close for the reasons you state. Now, what does any of this 'dropping a rock from a height' math have to do with rockets? $\endgroup$ Aug 28, 2022 at 3:39
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    $\begingroup$ Here is a link to the math formatting tutorial if you want to play with it math.meta.stackexchange.com/questions/5020/… $\endgroup$ Aug 28, 2022 at 15:29

1 Answer 1


It sounds like with ignoring drag and everything, this is simply showing what happens if the potential energy difference between the surface and 160km is transformed into kinetic energy.

$$U = -\frac{GMm}{r}$$ $$\Delta U = -GMm \left ( \frac 1 {R_{\text{earth}}+ 160{\text{km}}} - \frac 1 {R_{\text{earth} }}\right )$$ $$\Delta U = m (1.53 \times 10^6 \text{m}^2 \text{s}^{-2}) = \frac 1 2 m v^2$$ $$ v = \sqrt{2(1.53 \times 10^6 \text{m}^2 \text{s}^{-2})}$$ $$ v = 1750 \text{m/s}$$

Wolfram Alpha calculation

I would suspect the 1500m/s figure you've found represents some other quantity than the energy difference between surface and 160km.

  • $\begingroup$ So then, does this confirm my suspicion that I have been using a flawed logic? (Keep in mind, i'm an inexperienced high school freshman). I inferred that if an object would achieve a velocity of, let's say, -1750 m/s if dropped from some altitude, it would also take a rocket that can achieve a velocity of 1750 m/s to get that height (because, a rocket is simply fighting the potential energy of gravity with its own kinetic energy). However, i've always had a suspicion that this is a flawed logic. $\endgroup$
    – user48194
    Aug 28, 2022 at 15:15
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    $\begingroup$ @user48194 If you time-reverse the trajectory of a dropped object, you get the trajectory of a cannon shot, not a rocket - the velocity is highest at launch and drops off until it reaches zero at max height. Whereas if you shot a sounding rocket to 160 km, the velocity would be zero at launch, build to some max value at cutoff, then the rocket would coast to apogee at 160 km. Totally different trajectory. $\endgroup$ Aug 28, 2022 at 15:19
  • $\begingroup$ Oh... I finally understand. Thank you very much. So, what can I do if I wanted to figure out the velocity that a rocket would need to achieve to reach an apogee of 160 km? $\endgroup$
    – user48194
    Aug 28, 2022 at 15:30
  • $\begingroup$ @user48194 You can use your method to calculate the velocity needed at cutoff to coast to 160 km, but you will have to pick a cutoff altitude. If the cutoff altitude is zero, you get the cannon shot case. $\endgroup$ Aug 28, 2022 at 15:59
  • $\begingroup$ I see, so, could I use a piece wise function where the velocity starts dropping after cutoff altitude? $\endgroup$
    – user48194
    Aug 28, 2022 at 16:23

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