Showing posts with label gravitational forces. Show all posts
Showing posts with label gravitational forces. Show all posts

Obtain the value of $'g'$ from the motion of moon assuming that its period of rotation around the earth is $27$ days $8$ hours and the radius of its orbit is $60.1$ times the radius of the earth.

Given,

T = 27 days 8 hours = (27 * 24 + 8) hours = 656 hours = 2361600 Sec
Radius of Earth (R) = $6.36 * 10^6\;m$
Radius of orbit $(r) = 60.1 * R = 60.1 * 6.36 * 10^6$
Value of $g$ = ?

We have,
$T = $ $ \frac{2\;\pi\;r}{R}  \sqrt{ \frac{r}{g}}$

By Solving,
$g$ = $\frac{4\; \pi^2\;r^3}{T^2 \; R^2}$ = $\frac{4 \; \pi^2 \; * \; (60.1)^3 \; R^3}{T^2 \; R^2}$ = $\frac{4 \; \pi^2 \; * \; (60.1)^3 \; * \; 6.36\; * 10^6}{2361600}$ = $9.77 \; m/s^2$

∴ The required value of $g$ from the motion of the moon is $9.8\;m/s^2$
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The period of the moon revolving under the gravitational force of the earth is $27.3$ days. Find the distance of the moon from the center of the earth if the mass of the earth is $5.97 \;*\;10^{24}kg$.


Given,
Time $(T) = 27.3$ days = $27.3 \; * \; 24 \; * \; 60 \; * \; 60  = 2358720 \; Sec$
Mass of earth $(M) = 5.97 \; *\; 10^{24}\; kg$
Height of the Moon from the center of the earth $(h) = \;?$

We know that,
$h = $ $\left ( \frac{T^2 \; * \; R^2\; * \;g}{4\; \pi^2} \right )^{1/3}$ $ - \;R$
$h =$ $ \left ( \frac{(2358720)^2 \; * \; {6400000}^2\; * \; 9.8}{4\; \pi^2} \right )^{1/3}$ $ - \; 6400000 = 3.77 * 10^8 \; m$

∴ Height of the Moon from the center of the earth $(h) = \;3.77 * 10^8 \; m$

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Calculate the points along a line joining the centers of earth and moon where there is no gravitational force. ($M_e = 6 * 10^{24}\;kg, M_m = 7.4 * 10^{22}\;kg, d = 3.8 * 10^8\;m$).

Given, (need figure)

Mass of the earth $(M_e) = 6 * 10^{24}\;kg$
Mass of the Moon $(M_m) = 7.4 * 10^{22}\;kg$
Distance between centers of earth and the moon $(d) = 3.8 * 10^8\;m$

Let $P$ be the point at which gravitational force between earth and the moon is zero, then

$\frac{G\;*\;M_e\;*\;1}{x^2} = \frac{G\;*\;M_m\;*\;1}{(d-x)^2}$
$\frac{6\;*\;10^{24}}{x^2} = \frac{7.4\;*\;10^{22}}{(d\;-\;x)^2}$
$\frac{600}{x^2} = \frac{7.4}{(d-x)^2}$

$\left ( \frac{24.5}{x} \right )^2 = \left ( \frac{2.72}{d-x} \right )^2$
$\frac{24.5}{x} = \frac{2.72}{(d-x)}$
$27.22\;*\;x = 24.5\;*\;d$
$x = \frac{24.5\;*\;3.8\;*\;10^8}{27.22} = 3.42\;*\;10^8\;m$

∴ The distance from the earth center is $3.42\;*\;10^8\;m$

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An earth satellite moves in a circular orbit with a speed of $6.2\;km/s$. Find the time of one revolution and its centripetal acceleration.

Given,

Orbital velocity $v_o = 6.2\;km/s = 6200\;m/s$
Radius of earth $(R) = 6400000\;m$
Time Period $(T) = \; ?$
Centripetal Acceleration $(a) = \;?$

We have,
$ v_{0} = R \sqrt { \frac{g}{R+h}} $
or, $6200 = 6400000\;*\; \sqrt{\frac{10}{6400000 \; + \; h}}$
or, $9.38 \; *\; 10^{-7} = \sqrt{\frac{10}{6400000 + h}}$
or, $h = 10655301 - 6400000$
h = $4255301\;m$

Now, 
$T = 2 \pi \; \sqrt{\frac{(R + h)^3}{g * R^2}}$
or, $2\pi \; \sqrt{\frac{(4255301 + 6400000)^3}{10 * 6400000}}$
or, $2 \; \pi \;*\;1718.57 = 10792.65\;Sec = 2.99\;hrs$
∴ Time of revolution = $2.99\;hrs$

Again,
Centripetal Acceleration $a = \frac{v_0^2}{h + R} = \frac{(6200)^2}{4255301 + 6400000} = 3.6\;m/s^2$
∴ Centripetal Acceleration = $3.6\;m/s^2$

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Taking the earth to be the uniform sphere of radius $6400 \; km$, and the value of $g$ at the surface of earth $10\;m/s^2$, calculate the total energy needed to raise a satellite of mass $2000 \; kg$ to a height of $800 \; km$ above the ground and to set it into a circular orbit at that altitude.


Given;
Radius of earth (R) = 6400 km = 6400000 m
Mass of satellite (m) = 2000 kg
Height of satellite (h) = 800 km = 800000 m
Total Energy needed (E) = ?

We have,
Energy needed = Increase in Potential Energy + Kinetic Energy at Orbit

$\left [ -\;\frac{G\;M\;m}{r}\;-\;\left (\frac{-\;G\;M\;m}{R}\right )\; \right ]$

= $-$$\frac{G\;M\;m}{r}$$\;-\;$$\frac{-\;G\;M\;m}{R}$$\;+\;$$\frac{1}{2}$$mv^2$
= $-$$\frac{G\;M\;m}{r}$$\;-\;$$\frac{-\;G\;M\;m}{R}$$\;+\;$$\frac{1}{2}$$m$$ \frac{G\;M}{r}$                               [∵ r = R+h]

= $g\;R^2\;m[$$\frac{1}{R}$$\;-\;$$ \frac{1}{2(R+h)}$$]$     =     $g\;m[R\;-\;$$\frac{R^2}{2(R+h)}$$]$
= $2000\;*\;10\;[6400000 \;-\; $$\frac{(6400000)^2}{2(6400000 \;+\; 800000)}]$
= $7.12\;*\;10^{10}\;J$

∴ The total Energy needed (E) = $7.12\;*\;10^{10}\;J$

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An alpha particle is a nucleus of doubly ionized helium. It has mass of $6.68 * 10^{-27}$kg and charge of $3.2 * 10^{-19}C$. Compare the force of electrostatic repulsion between the two alpha particles with the force of gravitational attraction between them.

Given,

Mass of the alpha particle = $6.68\;*\;10^{-27}\;kg$
Charge of the alpha particle = $3.2 * 10^{-19}\;C$
We have,
Electrostatic Force $(E_f) = $$\frac{1}{4\pi \epsilon_0} \frac{q_1\;q_2}{R^2}$
Electrostatic Force $(E_f) = 9*10^9 * $$\frac{q^2}{R^2}$ = $9*10^9 * $$\frac{(3.2 * 10^{-19})^2}{R^2}$ ......... (i)

Gravitational force $(G_f) = G $$\frac{M_1\;M_2}{R^2}$
Gravitational force $(G_f) = 6.67\;*\;10^{-11} $$\frac{(6.68*10^{-27})^2}{R^2}$ .......... (ii)

Dividing equation (i) and (ii), we get
$\frac{E_f}{G_f} = \frac{9.216 \;*\; 10^{-28}}{2.976 \;*\; 10^{-63}}$
$= 3.096 \;*\; 10^{35}$
⇒ $E_f = 3.096 \;*\; 10^{35}\; G_f$

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Theory of Gravity - General Relativity


“Einstein, my upset stomach hates your theory [of General Relativity] - it almost hates you yourself! How am I to' provide for my students? What am I to answer to the philosophers?” 
~ Paul Ehrenfest

Let’s talk about the birth of our most successful theory of gravity - general relativity.


Prior to Einstein, it was Newton’s Law of Universal Gravitation that was the accepted theory of gravity. All of the gravitational phenomena in the Universe, from the acceleration of masses on Earth to the orbits of the moons around the planets to the planets themselves revolving around the Sun, his theory described it all. 

Objects exerted equal-and-opposite gravitational forces on one another, they accelerated in inverse proportion to their mass, and the force obeyed an inverse square law. By the time the 1900s rolled around, it had been incredibly well-tested, and there were no exceptions. Well, with thousands upon thousands of successes to its credit, there were almost no exceptions, at any rate.


In the Newtonian picture of gravity, space and time are absolute, fixed quantities, while in the Einsteinian picture, spacetime is a single, unified structure where the three dimensions of space and the one dimension of time are inextricably linked.

The first major development was that space and time, previously treated as a separate three-dimensional space and a linear quantity of time, were united in a mathematical framework that created a four dimensional “spacetime.” This was accomplished in 1907 by Hermann Minkowski:

The views of space and time which I wish to lay before you have sprung from the soil of experimental physics, and therein lies their strength. […] Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality.

This worked only for flat, Euclidean space, but the idea was incredibly powerful mathematically, as it led to all the laws of special relativity as an inevitable consequence. When this idea of spacetime was applied to the problem of Mercury’s orbit, the Newtonian prediction under this new framework came a little closer to the observed value, but still fell short.

Minkowski’s spacetime corresponded to an empty Universe, or a Universe with no energy or matter of any type.


Countless scientific tests of Einstein’s general theory of relativity have been performed, subjecting the idea to some of the most stringent constraints ever obtained by humanity. Einstein’s first solution was for the weak-field limit around a single mass, like the Sun; he applied these results to our Solar System with dramatic success. (LIGO scientific collaboration / T. Pyle / Caltech / MIT)

Einstein was able to find a solution where you had a Universe with one single, solitary point mass source in it, and with the stipulation that you were outside of that point. This reduced to the Newtonian prediction at great distances, but gave stronger results at closer distances. 

These results not only agreed with the observations of Mercury’s orbit that Newtonian gravity failed to predict, but it made new predictions about the deflection of starlight that would be visible during a total solar eclipse, predictions that were later confirmed during the solar eclipse of 1919.

This article is adapted from the  medium.com. Click here for more details.