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Gravitational Force (Numerical Problem)

Q. If mass of the sun is 2 × 1030 kg, that of the earth is 6 × 1024 kg and the distance between them is 1.5 × 1011 m. What is the gravitational force produced between them? Where, G = 6.67× 10− 11 N m2/kg2 Soln→ Mass of sun m1 = 2 × 1030 kg Mass of earth m2 = 6 × 1024 kg distance between them d = 1.5 × 1011 m. Gravitational force between them F = ? By the formula, F = G $\frac{\mathrm{m<sub>1</sub>\; m<sub>2</sub>}}{\mathrm{d\; <sup>2</sup>}}$ = 6.67 × 10− 11 $\frac{\mathrm{2\; \times \; 10<sup>30</sup>\; 6\; \times \; 10<sup>24</sup>}}{\mathrm{(1.5\; \times \; 10<sup>11</sup>\; )<sup>2</sup>}}$ =	Q. Calculate the force of gravitational due to the earth on a boy 20 kg standing on the surface of the earth. [mass of the earth 6 × 1024 kg and the radius is 6.4 × 10 3km.] Soln→ Mass of earth m1 = 6 × 1024 kg Mass of boy m2 = 20kg Radius of Earth R = 6.4 × 103 m. Gravitational force between them F = ? By the formula, F = G $\frac{\mathrm{m<sub>1</sub>\; m<sub>2</sub>}}{\mathrm{R\; <sup>2</sup>}}$ = 6.67 × 10− 11 $\frac{\mathrm{6\; \times \; 10<sup>24</sup>\; \times \; 20}}{\mathrm{(6.4\; \times \; 10<sup>3</sup>)<sup>2</sup>}}$ =
Q. The gravitational force produced between any two object kept 2.5 × 104 km apart is 500N. At what distance should they be kept so that the gravitational force becomes half? Soln→ Gravitational force between two object F = 500 N. Distance between is d1 = 2.5 × 104 km = 2.5 × 104 × 1000 m = 2.5 × 107 m Now, Gravitational force F2 = $\frac{F}{2}$ = $\frac{\mathrm{500}}{2}$ = 250N The new distance between them d2 = ? $\frac{\mathrm{F<sub>1</sub>}}{F}$ = $\frac{\mathrm{(d<sub>1</sub>)<sup>2</sup>}}{\mathrm{d<sup>2</sup>}}$ $\frac{\mathrm{250}}{\mathrm{500}}$ = $\frac{\mathrm{(d<sub>1</sub>)<sup>2</sup>}}{\mathrm{(2.5\; \times \; 10<sup>7</sup>)<sup>2</sup>}}$ (d1)2 = $\frac{\mathrm{250}}{\mathrm{500}}$ × (2.5 × 107)2 (d1)2 = 0.5 × 6.25 × 1014 (d1)2 = 3.125 × 1014 d1 = 1.77 × 107 m.	Q. The mass of the moon is 7 × 1022 kg and the radius is 1.7 × 10 6m, what will be the gravitational acceleration of the moon? What will be the weight of a man of 70 kg mass on the moon? Soln→ Mass of moon M = 7 × 1022 kg Radius of moon R = 1.7 × 10 6m Gravitational acceleration of moon g = ? By the formula, F = G $\frac{M}{\mathrm{R\; <sup>2</sup>}}$ = 6.67 × 10− 11 $\frac{\mathrm{7\; \times \; 10<sup>22</sup>}}{\mathrm{(1.7\; \times \; 10<sup>6</sup>)<sup>2</sup>}}$ =
Q. The radius of the earth is 6.37 × 103 km and height of Mt. Everest from the sea level is 8848m, if the value of 'g' on the surface of the earth is 9.8 m/sec2, Calculate the valueof 'g' at the top of Mt. Everest? Soln→ Radius of earth R = 6.37 × 103 km = 6.37 × 103 × 1000 m = 6.37 × 106 m Height of Mt. Everst h = 8848 m Value of acceleration g on earth surface = 9.8 m/sec2 Value of acceleration g at top of the Mt. Everst g1 = ? By the formula, $\frac{\mathrm{g<sup>1</sup>}}{g}$ = ($\frac{R}{\mathrm{(R+h)}}$)2 g1 = 9.8 × ($\frac{\mathrm{6.37\; \times \; 10<sup>6</sup>}}{\mathrm{6.37\; \times \; 10<sup>6</sup>+8848}}$)2 g1 = 9.8 × ($\frac{\mathrm{6.37\; \times \; 1000000}}{\mathrm{(6370000+8848)}}$)2 g1 = 9.8 × ($\frac{\mathrm{6370000}}{\mathrm{(6378848)}}$)2 g1 = 9.8 × 0.992 g1 = 9.8 × 0.9801 g1 = 9.6 m/sec2 Q. Calculate the acceleration of a meteor located at the height of 925 km from the surface of the earth. The mass and radius of the earth is 6 × 1024 kg and 6400 km respectively.	Q. The gravity of the earth is six times greater than that of the moon. How much kg mass can a person lift on the moon if he can lift 80kg mass on the earth? Soln→ Maximum force a man can apply on earth surface wE = Maximum force a man can apply on moon surface Wm ME g E = Mm gm 80 × 9.8 = Mm × 1.67 784 = Mm × 1.67 Mm = $\frac{\mathrm{784}}{\mathrm{1.67}}$ Mm = 469.46 kg. Hence, that man can lift 469.46kg on moon surface. Q. Two ball of equal mass are kept at a distance of 100m from their centers. If the gravitational force between them is 1575N, calculate the mass of each sphere. Q. The mass of the earth is 6 × 1024kg and its radius is 6400 km. Calculate the weight of an object of mass 250kg on the surface of the earth.
Q. Mass of earth 6 × 1024kg and radius 6400 km. A person weight on spring balance is 977N then what is the mass of that man? Soln→ Mass of earth M = 6 × 1024kg Radius of earth R = 6400km = 64000 × 1000 = 6400000 m =6.4 × 106 Weight of man w = 977 N By the formula, w = G $\frac{\mathrm{M\; m}}{\mathrm{R\; <sup>2</sup>}}$ m = $\frac{\mathrm{F\; \times \; R<sup>2</sup>}}{\mathrm{G\; M}}$ = $\frac{\mathrm{977\; \times \; (6.4\; \times \; 10<sup>6</sup>)<sup>2</sup>}}{\mathrm{6.67\; \times \; 10<sup>-\; 11</sup>}}$ =$\frac{\mathrm{40017.92\; \times \; 10<sup>12</sup>}}{\mathrm{40.02\; \times \; 10<sup>13</sup>}}$ = 999.94 × 1012 − 13 = 999.94 × 10 − 1 = 99.99 kg	Q. Two masses 'x' kg and 200kg are 20m apart. Gravitational force between them is 3.335 × 10 − 9 N, then find the value of 'x'? Soln→ Mass of first object m1 = x kg Mass of the second object m2 = 200kg Distance between them d = 20 m Gravitational force F = 3.335 × 10 − 9 N By the formula, w = G $\frac{\mathrm{m<sub>1</sub>\; m<sub>2</sub>}}{\mathrm{d\; <sup>2</sup>}}$ m1 = $\frac{\mathrm{F\; \times \; d<sup>2</sup>}}{\mathrm{G\; m<sub>2</sub>}}$ x = $\frac{\mathrm{3.335\; \times \; 10\; <sup>\; -\; 9\; </sup>\; \times \; (20)<sup>2</sup>}}{\mathrm{6.67\; \times \; 10<sup>-\; 11</sup>\; \times \; 200}}$ =$\frac{\mathrm{3.335\; \times \; 400\; \times \; 10<sup>-\; 9</sup>}}{\mathrm{6.67\; \times \; 200\; \times \; 10<sup>-\; 11</sup>}}$ = 1 × 10 − 9 + 11 = 1 × 10 2 = 100 kg