Question

the mass of the earth is 6×10^24 kg and its radius is 6400 km.calculate the height above the earth's surface at which value of acceleration due to gravity will be 4m/s2.​

Answer 1

$\huge \bold \color{khaki}{ \underline \color{indigo} \mathfrak{Answer }}$

$g = \frac{GM_e}{ {R_e}^{2} }$

$where \\ g = acceleration \: due \: to \: gravity \\ G = gravitational \: constant \\ M_e = mass \: of \: earth \\ R_e = radius \: of \: earth$

$g = \frac{6.67 \times {10}^{ - 11 \times} \times 6 \times {10}^{24} }{( {6 \times {10}^{5} )}^{2} } = 9.8m/s$

When the object is taken to a height h from the surface of earth,

${g}^{ \circ} = \frac{gm_e}{ {(R_e + h)}^{2} }$

${g}^{ \circ} = \frac{GM_e}{ {(R_e(1 + \frac{h}{R_e})) }^{2} }$

${g}^{ \circ} = \frac{g}{( {1 + \frac{h}{R_e}) }^{2} }$

${g}^{ \circ} = g( {1 + \frac{h}{R_e} })^{ - 2}$

Applying Binomial Expression,

${1 + x}^{n} = 1 + nx \: if \: x < < 1$

$\because \: h < < R_e \\ \frac{h}{R_e} < < 1$

$\therefore \: {g}^{ \circ} = g(1 - \frac{2h}{R_e} )$

Putting g°= 4 m/s^2, g= 9.8 m/s^2 and Re = 6400000m,

$4 = 9.8(1 - \frac{2h}{6400000} )$

$2h = 0.74\times {10}^{ - 3}$

$\color{plum} \boxed { \bold\color{darkgreen} h = 0.37 \times {10}^{ - 3} m}$

$\mathbb{Hope \: it \: Helps... }$