A body weighs 72 N on the surface of the earth- What is the gravitational force on it- a height equal to half the radius of the earth -

Question

Toppr Admin · Accepted Answer

Correct option is A- -32- N -W-s- - mg-s- - 72- N -160-W-h- - mg-h- - -dfrac-mg-s-left -1 - -dfrac-h-R- -right -2- - -dfrac-72- N -left -1 - -dfrac-R-2-R- -right -2- - -dfrac-72-9-4-160- -W-h- - 32- N -160-

A body weighs $$72\ N$$ on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth ?

Correct option is A. $$32\ N $$
$$W_{s} = mg_{s} = 72\ N $$

$$W_{h} = mg_{h} = \dfrac{mg_{s}}{\left (1 + \dfrac{h}{R} \right )^{2}} = \dfrac{72\ N }{\left (1 + \dfrac{R/2}{R} \right )^{2}} = \dfrac{72}{9/4} $$

$$W_{h} = 32\ N $$

A body weighs $$72\ N$$ on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth ?

Correct option is A. $$32\ N $$$$W_{s} = mg_{s} = 72\ N $$ $$W_{h} = mg_{h} = \dfrac{mg_{s}}{\left (1 + \dfrac{h}{R} \right )^{2}} = \dfrac{72\ N }{\left (1 + \dfrac{R/2}{R} \right )^{2}} = \dfrac{72}{9/4} $$$$W_{h} = 32\ N $$

Correct option is A. $$32\ N $$
$$W_{s} = mg_{s} = 72\ N $$

$$W_{h} = mg_{h} = \dfrac{mg_{s}}{\left (1 + \dfrac{h}{R} \right )^{2}} = \dfrac{72\ N }{\left (1 + \dfrac{R/2}{R} \right )^{2}} = \dfrac{72}{9/4} $$

$$W_{h} = 32\ N $$