A body weighs 63 N on the surface of the earth. What is the gravitational force (in newton) on it due to the earth at a height equal to half the radius of the earth?

Question

A body weighs 63 N on the surface of the earth. What is the gravitational force (in newton) on it due to the earth at a height equal to half the radius of the earth? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Here, m g=63 N , h=R / 2 As (gh/g)=((R/R+h))2=((R/R+(R)2))2 =((2/3))2=(4/9) gh=(4/9) g ∴ m gh=(4/9) m g =(4/9) × 63=28 N

Q. A body weighs $63 N$ on the surface of the earth. What is the gravitational force (in newton) on it due to the earth at a height equal to half the radius of the earth?

Here, $m g = 63 N,$
$h = R /2$
As $\frac{g _{h}}{g} = (\frac{R}{R + h})^{2} = (\frac{R}{R + \frac{R}{2}})^{2}$
$= (\frac{2}{3})^{2} = \frac{4}{9}$
$g_{h} = \frac{4}{9} g$
$∴ m g_{h} = \frac{4}{9} m g$
$= \frac{4}{9} \times 63 = 28 N$

Q. A body weighs 63N on the surface of the earth. What is the gravitational force (in newton) on it due to the earth at a height equal to half the radius of the earth?

Here, mg=63N, h=R/2 As ggh​​=(R+hR​)2=(R+2R​R​)2 =(32​)2=94​ gh​=94​g ∴mgh​=94​mg =94​×63=28N

Q. A body weighs $63 N$ on the surface of the earth. What is the gravitational force (in newton) on it due to the earth at a height equal to half the radius of the earth?

Here, $m g = 63 N,$
$h = R /2$
As $\frac{g _{h}}{g} = (\frac{R}{R + h})^{2} = (\frac{R}{R + \frac{R}{2}})^{2}$
$= (\frac{2}{3})^{2} = \frac{4}{9}$
$g_{h} = \frac{4}{9} g$
$∴ m g_{h} = \frac{4}{9} m g$
$= \frac{4}{9} \times 63 = 28 N$