Physics, asked by Anonymous, 9 months ago

Derive the concept of gravity.
No copy or spam or improper answer is allowed is allowed!!
Any sort of violation will be reported !!
So, be careful!!

Answers

Answered by anishka56
5

&lt;font color="red"&gt;[tex]<b

✒✒F=Gm1 m2 /r2. Another, common, Gravity formula is the one you learned in school; the acceleration due to the gravity of the Earth , on a test mass. This is, by convection, written as g, and is easily derived from the gravity formula above ( M is the mass of the earth, and r its radius) : GM/ r2✔✔

⤴️⤴️please mark me as a brainleist and follow me plz.

Answered by Anonymous
4

A knowledge of acceleration due to gravity enables us to determine the following three quantities, namely_

(i) Mass of the earth

(i) Mass of the earth(ii) Density of the earth

(i) Mass of the earth(ii) Density of the earth(iii) Mass of the sun

Now let's come to the point of derivation_

(i) Mass of the earth:

                                Mass 'M' of the earth can be calculated fro the equation (9) of universal gravitational constant (G).

Such that,

M = gR^2/g

since g = 9.8 m/s^2

R= 6.38*10^6

G= 6.67*10 ^11

∵ Metric system

M=9.8 * 10^6/6.67 * 10^-11

=> 5.98 * 10^-11

So, the mass of the earth is 5.98*10²⁴ kg.

(ii) Density of the earth:

                                   If ' ρ' is the mean density of the earth then,

M= V * I

= 4/3) IGRP

                             

Substituting for M in equation (9),

I = 3g/4IGRP

=>5497kg/mA^3

So, the density of the earth is 5497 kg/m³

(iii) Mass of the sun:

                               If 'm' be the mass of the sun, gravitational attraction between the sun, and the earth is_

Where 'r' is the radius of orbit around the sun and M is the mass  of the earth. Centripital force 'Fc' required by the earth is_

Where, 'ω'  is the angular velocity of the earth.

In equlibrium,

In equlibrium,Fc= Fg

In equlibrium,Fc= FgmrIa^2= GMm/rA^2

Where 'T' is the time period of rotation of earth around the sun.

Here,

r = 150 million km

r = 150 million km = 150*10 ⁶ km

r = 150 million km = 150*10 ⁶ kmor, r = 1.5*10⁸ km

r = 150 million km = 150*10 ⁶ kmor, r = 1.5*10⁸ km = 1.5*10¹¹ km

r = 150 million km = 150*10 ⁶ kmor, r = 1.5*10⁸ km = 1.5*10¹¹ kmG = 6.67*10⁻¹¹ M.K.S. units

r = 150 million km = 150*10 ⁶ kmor, r = 1.5*10⁸ km = 1.5*10¹¹ kmG = 6.67*10⁻¹¹ M.K.S. units = 365*24*60*60s

So, the mass of the sun is 2*10³⁰ kg.

Similar questions