Acceleration due to gravity is -g- on the surface of the earth- The value of acceleration due to gravity a height of 32 km above earth-s surface is -Radius of the earth -6400 km-

Question

Acceleration due to gravity is -g- on the surface of the earth- The value of acceleration due to gravity a height of 32  km above earth-s surface is -Radius of the earth -6400  km-

Toppr Admin · Accepted Answer

Correct option is B- -0-99 - g-h- 32 - km- R-6400 - km- - so -h-lt-lt-R-g- g-bigg- 1- -dfrac-2h-R- -bigg- - g -bigg-1- -dfrac-2 -times 32-6400- -bigg- -implies g- -dfrac-99-100- g- 0-99 - g-

Acceleration due to gravity is '$$g$$' on the surface of the earth. The value of acceleration due to gravity at a height of $$32 \ km$$ above earth's surface is (Radius of the earth $$=6400 \ km$$)

Correct option is B. $$0.99 \ g$$
$$h= 32 \ km, R=6400 \ km, $$ so $$h<<R$$

$$g'= g\bigg( 1- \dfrac{2h}{R} \bigg) = g \bigg(1- \dfrac{2 \times 32}{6400} \bigg) $$

$$\implies g'= \dfrac{99}{100} g= 0.99 \ g$$

Acceleration due to gravity is '$$g$$' on the surface of the earth. The value of acceleration due to gravity at a height of $$32 \ km$$ above earth's surface is (Radius of the earth $$=6400 \ km$$)

Correct option is B. $$0.99 \ g$$$$h= 32 \ km, R=6400 \ km, $$ so $$h<<R$$$$g'= g\bigg( 1- \dfrac{2h}{R} \bigg) = g \bigg(1- \dfrac{2 \times 32}{6400} \bigg) $$$$\implies g'= \dfrac{99}{100} g= 0.99 \ g$$

Correct option is B. $$0.99 \ g$$
$$h= 32 \ km, R=6400 \ km, $$ so $$h<<R$$

$$g'= g\bigg( 1- \dfrac{2h}{R} \bigg) = g \bigg(1- \dfrac{2 \times 32}{6400} \bigg) $$

$$\implies g'= \dfrac{99}{100} g= 0.99 \ g$$