Calculation of gravitational force:-

Suddenly it came to my mind on 19th November night that the current calculation of gravitational force is not correct and calculation should be divided into two parts, actual and real gravitational force. Please consider my below findings.

Currently we are calculating the gravitational force of a planet/moon like below:-

According to Newton gravitational formula we calculate like below:-

Fgrav = (Gm1m2)/d2

• Fgrav is the force due to gravity • G is the universal gravitation constant 6.673 x 10-11 Nm2/kg2[4] • m1 is the mass of the first object • m2 is the mass of the second object • d is the distance between the centers of two objects

But here we never consider one very important factor i.e. the centrifugal force of the planet that has rotation. Due to the rotational speed the centrifugal force of the planet will increase and it will reduce the gravitational force of the planet.

For example the current acceleration due to gravity on earth surface is 9.8m/s2 and current rotational speed is 1670 kilometers/hour. But if this rotational speed would have been 3340 kilometers/hour the acceleration due to gravity on earth surface would have been very less than current one. So current gravitational force is not actual gravitational force of earth. If a planet with the size of Jupiter with proper amount of water and oxygen still we cannot even imagine that a human can live on that planet due to very high level of gravitational force ( for example a human being having 75kg of weight would become 6,375 kg on Jupiter surface and in such a situation a human being will not be able to even stand). But if the rotational speed of that planet is so high( for example 141950 Kilometers/hour) then the acting gravitational force become nearer to 9.8m/s2 and with that gravity human being can live even in a very larger planet based on its rotational speed which will decrease the gravitational force.

I might not be 100% correct in the calculation but real gravitation force of a planet is relates to its centrifugal force due to rational speed is like below:-

Fgrav ∝ 1/ CF (centrifugal force)

If centrifugal force increases real gravitational force will decrease and vice versa.

For example if earth’s rotational speed increase from 1670 kilometers/hour to 3340 kilometers/hour weight of all the particle on earth surface become ½ of its original weight and one full day duration will become 12 hours instead of 24 hours.


closed as unclear what you're asking by dmckee Dec 28 '17 at 18:52

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ Generally, this is not a place to propose new ideas or have them reviewed by others. Also note that physicists are quite aware of centrifugal force due to the rotating planet. I'm not sure what this is supposed to resolve; since I don't see anything revolutionary here. $\endgroup$ – JMac Dec 28 '17 at 17:25
  • $\begingroup$ I know every one know centrifugal force due to the rotation of planet. But as per my knowledge we never considered the rotational speed of the planet in gravitational force calculation. $\endgroup$ – Abhijit Pritam Dutta Dec 28 '17 at 17:29
  • $\begingroup$ Because in gravitational force calculation you are calculating the gravitational force; which in Newtonian mechanics is defined through the formula you've given. It's entirely proportional to the masses and distance between them. The centrifugal force is an additional force that acts in approximately the same directions. It doesn't mean it should be considered the same force. The added effect of centrifugal force is relatively small compared to gravity; and gravitational acceleration ($g$) may already factor this in (depending on how precise it is in the first place). $\endgroup$ – JMac Dec 28 '17 at 17:36
  • $\begingroup$ Oh, it is a lot more complex than this. Consider frame dragging which you forgot to account for. $\endgroup$ – ja72 Dec 28 '17 at 18:33
  • $\begingroup$ Jupiter also doesn't exactly have a surface. It just gets denser as you go deeper. $\endgroup$ – zeta-band Dec 28 '17 at 19:17

You haven't came up with a new theory. You are just confusing gravity, centrifugal force and weight.

Imagine a person of mass $m$ standing on earth. There are only two forces acting on him

  • The earth is pulling him with a force $F=\frac{GM_{\rm E}m}{R_{\rm E}^{2}}$ downwards.
  • The ground pushes him upwards to keep him from falling down with a force $N$.

Because the person rotates with earth at $\omega_{\rm E}=\frac{2\pi}{T_{\rm E}}$ ($T_{\rm E}=24\:\rm Hours$), he accelerate with acceleration $a=m\omega_{\rm E}^{2}R_{\rm E}$ (this is true only at the equatorial) towards the center of earth. Newton tells us that

$$\Sigma F=ma$$

$$\frac{GM_{\rm E}m}{R_{\rm E}^{2}}-N=m\omega_{\rm E}^{2}R_{\rm E}$$

$$N=mg-m\omega_{\rm E}^{2}R_{\rm E}=m\left(g-\omega_{\rm E}^{2}R_{\rm E}\right)$$

with $g\equiv\frac{GM_{\rm E}}{R_{\rm E}^{2}}\approx9.8\:\rm m/sec^{2}$. The normal force $N$ is what you perceive as weight, not the gravitational force. This is what your scale will read when you stand on them. You can clearly see that as you spin faster - you weigh less. This is the centrifugal effect you mentioned, but you see that when calculating things it is taken as an additional effect other than gravity. But if you'll plug the numbers you'll see that

$$\omega_{\rm E}^{2}R_{\rm E}\approx0.03\:\rm m/sec^{2}\ll g$$

which means that the effect of the earth rotation is negligible compared to gravity.

EDIT: As @JMac and @dmckee mentioned in the comments, since the earth is not a perfect sphere the gravitational acceleration $g$ has a spatial dependency and is differ from place to place on earth


So if you want to generalize the above expression to every location on earth you'll have

$$N\left(\theta,\varphi\right)=m\Big(g\left(\theta,\varphi\right)-\omega_{\rm E}^{2}R_{\rm E}\left(\theta,\varphi\right)\Big)$$

where $\theta$ is the azimuth angle measured from the rotation axis. In addition, following @dmckee, the variation of $g$ with $\left(\theta,\varphi\right)$ is of the same order of magnitude as the centrifugal part

$$\Delta g\sim \omega_{\rm E}^{2}R_{\rm E}$$

This means that it is meaningless to treat $g$ as a constant and also account for the centrifugal effect. The bottom line is that it is best to define an effective gravitational acceleration

$$g_{\rm eff}\left(\theta,\varphi\right)\equiv g\left(\theta,\varphi\right)-\omega_{\rm E}^{2}R_{\rm E}\left(\theta,\varphi\right)$$

and just measure it experimentally at different locations on earth. Then your weight is just

$$N=mg_{\rm eff}$$

with the suitable $g_{\rm eff}$ for your area.

  • $\begingroup$ Thanks eranreches. I have done my science till 12th only and hence probably my thought might not be correct. Are you saying based on above calculation that even if earth rotational speed increase to double of its current speed the gravitation force will still the same and a personal will remain with the same weight? $\endgroup$ – Abhijit Pritam Dutta Dec 28 '17 at 17:58
  • $\begingroup$ @Abhijit I am saying that the gravitational force will remain the same, but a person will weigh less due to an additional effect. $\endgroup$ – eranreches Dec 28 '17 at 18:00
  • $\begingroup$ OK thanks. That means there are existing formula where it consider the rotational speed while calculating the Wight of a thing in the surface of a planet? I just know weight = mass x gravity $\endgroup$ – Abhijit Pritam Dutta Dec 28 '17 at 18:05
  • 2
    $\begingroup$ Another reason to not be explicit about the centrifugal contribution to weight is that even along lines of constant latitude you will find that local $g$ varies at a scale that is comparable to or even larger than the equatorial centrifugal contribution due to local geography (a fact exploited by people search for water, oil, and minerals). $\endgroup$ – dmckee Dec 28 '17 at 18:54
  • 1
    $\begingroup$ Nice bit on the relationship between the centrifugal and non-sphericial effects: physics.stackexchange.com/questions/8074/…. The fact that the local geographical effect are also of that size seems to be a coincidence. $\endgroup$ – dmckee Dec 28 '17 at 20:12

Not the answer you're looking for? Browse other questions tagged or ask your own question.