Question

Question 36:
What is the force of gravity on a body of mass 150 kg lying on the surface of the earth? (Mass of earth = 6 x 1024 kg; Radius of earth = 6.4 x 106 m; G = 6.7 x 10-11 Nm2/kg2)

Lakhmir Singh Physics Class 9

Answer 1

Answer:

1471.5 N

Explanation:

Given,

$\textrm{Mass\ of\ earth,\ M}\ =\ 6\times 10^{24}\ Kg\\\\\textrm{Radius\ of\ earth, R}\ =\ 6.4\ \times10^{6}\ Km\\\\\textrm{Gravitation\ constant,\ G}\ =\ 6.7\times\ 10^{-11}\ \dfrac{Nm^{2}}{Kg^{2}}\\\\\textrm{Now,\ the\ acceleartion\ due\ to\ gravity\ is\ given\ by,}\ g\ =\ \dfrac{G\times M }{R^2}\\\\=\ \dfrac{6.7\times\ 10^{-11}\times 6\times 10^{24} }{(6.4\ \times10^{6})^{2}}\ =\ 9.81\ m/s$

Now, the given mass of body lying on the surface , m= 150 kg

So, the force acting on the body is given by, F= m.g

                                                                             = 150 x 9.81 N

                                                                             = 1471.5 N

So, the force of gravity on a body of mass 150 kg lying on the surface of the earth will be 1471.5 N.

Answer 2

Answer:

1472.1 N

EXPLANATION:

We have to calculate the force of gravity (F) on the body;

mass of the body = 150 kg

Mass of earth = 6 x 1024 kg;

Radius of earth = 6.4 x 106 m;

G = 6.7 x 10-11 Nm2/kg2)

Force of gravity is basically the force exerted on the body of mass m , due to the gravitational force of the Earth .

So, the force of gravity (F) should be equal to mass of the object multiplied by the mass of the earth and the gravitational of the earth, divided by the square of the radius of the eartyh.

∴ $F = \frac{G \times M \times m}{r^{2} }$

$F = \frac{6.7 \times 10^{-11 } \times 6 \times 10^{24} \times 150  }{(6.4 \times 10^{6)} ^{2} }$

∴$\frac{6030 \times  10^{13} }{40.96 \times 10^{12} }$

= 1472.1