Question

Mathematically express the acceleration due to gravity in terms of mass of earth and radius of earth

Answer 1

Answer:

Explanation:

Answer:how does the value of ‘g’ on earth related to mass of earth and radius. Derive it

Explanation:

Consider a body of mass ‘m’ held at a distance of ‘r’ from the center of the earth (radius of earth =R) of mass ‘M’

According to Newton’s law of gravitation:

F=Gm1m2/r^2

F=GmR/R^2. .........(1)

Acc. To Newton’s second law of motion

F=mg. .........(2)

From equations (1) and (2)

GmM/R^2=mg

(m and m cancels)

GM/R*2=g

6.67*10^-11*6*10^24

=9.77~ approx 9.8

Hence,g= 9.8m/s*2

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Answer 2

Acceleration due to gravity on earth is 9.8 m/s²

Explanation:

• Newton's Law of Gravitation states that between any two objects in the universe, a force of gravitation or attractive force is present which is directly proportional to the product of their individual masses and inversely proportional to the square of the distance between their centers.

• If the two objects are a random load and earth of mass $m_{1}$ and $m_{2}$respectively and the distance between them is $R$ , then,

          Mathematically,   $F=G\frac{m_{1} m_{2} }{R^{2} }$      

• From Newton's second law of motion, we know that the external force applied on any body is directly proportional to its respective change in momentum with respect to the time. If force is applied on mass $m_{1}$ , then

           Mathematically,   $F=m_{1}g$      

Equating both the value of forces, we get

$g=G\frac{m_{2} }{R^{2} }$

We know, value of

$G = 6.67$ × $10^{-11}Nm^{2}/ kg^{2}$

$m_{2}=6$ × $10^{24}kg$    

$R=6.3$ × $10^{6}m$  

Substituting the given values in the equation, we get

$g=\frac{(6.67*10^{-11} )(6*10^{24} )}{(6.3*10^{6} )^{2} }\\g=\frac{(6.6)(6)(10^{13} )}{(6.3)^{2}(10^{12} ) }$

$g=\frac{(66)(6)}{(6.3)(6.3)}$    $;$ $hence,$

$g=9.8m/s^{2}$  $(approx)$