Question

Find the binding energy of a body of mass 50kg at rest on the surface of the earth. [ Given : G = 6.67 x 10⁻¹¹ Nm² / kg² , R = 6400 km, M = 6 x 10²⁴ kg] (Ans : 3.127 x 10⁹ J)

Answer 1

Answer:

$\textrm{binding\ energy\ }=\ 3.127\times10^9$

Explanation:

Given,

mass of body = 50 kg

$\textrm{gravitation constant, G }= 6.67\times 10^{-11}\ Nm^2/kg^2$

$\textrm{Mass of earth}, M= 6\times 10^{24}\ kg$

$\textrm{radius of earth, R\ }=\ 6400\ km$

                              = 6400000 m

acceleration due to gravity,

                               $g\ =\ \dfrac{G.M}{R^2}$

                                    $=\dfrac{6.67\times 10^{-11}\times 6\times 10^{24}}{6400^2}$

                                     $= 9.77\ \dfrac{m}{s^2}$

So, the binding energy of the object = m.g.R

                                                            $=\ 50\times 9.77\times 6400000$

                                                              $= 3.127\times 10^9 J$

Hence, the binding energy of the body will be $3.127\times 10^9 J.$

Answer 2

Answer:

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