Question

At what height from the surface of earth the value of g is 6 ms–2 (R = 6400 km)​

Answer 1

Solution:

$g_{h} = Acceleration \ due\ to \ gravity \ from \ particular \ height = 6 \ ms^{-2}$

$g_{e} = Acceleration \ due\ to \ gravity \ on \ earth = 9.8 \ ms^{-2}$

$R = Radius \ of \ Earth = 6400 \ km = 6400 * 10^{3} m$

$h = Height \ from \ where \ Acceleration \ due\ to \ gravity \ is \ 6 \ ms^{-2} = \ h * 10^{3}$

$g_{h} = g_{e} [ \frac{R}{R + h} ]^{2}$

$6 = 9.8 [ \frac{6400 * 10^{3} }{(6400 + h )* 10^{3} }]^{2}$

$\frac{6}{9.8} = [ \frac{6400 * 10^{3} }{(6400 + h )* 10^{3} }]^{2}$

$\frac{6}{9.8} = [ \frac{6400}{(6400 + h )}]^{2}$

$\frac{6}{9.8} = \frac{40960000}{40960000 + h^{2} + 12800h }$

$40960000 + h^{2} + 12800h = \frac{40960000 * 9.8}{6 }$

$40960000 + h^{2} + 12800h = {40960000 * 1.6}$

$40960000 + h^{2} + 12800h = 65536000$

$h^{2} + 12800h = 65536000 - 40960000$

$h^{2} + 12800h = 24576000$

$h^{2} + 12800h - 24576000 = 0$

General form of Quadratic Equation => $ax^{2} + bx + c$

We have a quadratic equation let's take out roots.

$Discriminant = b^{2} - 4ac$

$Discriminant = 12800^{2} - 4 * 1 * 24576000$

$Discriminant = 262144000$

Solutions to Quadratic Formula:-

$h = \frac{-b \pm \sqrt{b^{2} - 4ac} }{2a}$

$h = \frac{-12800 \pm \sqrt{262144000} }{2 * 1}$

$h = \frac{-12800 \pm \sqrt{262144000} }{2}$

$h = \frac{-12800 + \sqrt{262144000} }{2} , h = \frac{-12800 - \sqrt{262144000} }{2}$

$h = \frac{-12800 + 5120\sqrt{10} }{2} , h = \frac{-12800 - 5120\sqrt{10} }{2}$

$h = -6400 + 2560\sqrt{10} , h = -6400 - 2560\sqrt{10}$

$h = -6400 + 8095.43 , h = -6400 - 8095.43$

$h = 1695.43 , h = -14495.43$

We will take positive value of h.

The height from the surface of earth will be 1695.43 m

Answer 2

At 1782.35 km km from the surface of earth the value of g is 6 m/s².

Explanation:

The acceleration due to gravity at different altitude is given by the formula:

gh/g = R²/(R + h)²

Where,

gh = Acceleration due to gravity at certain height = 6 m/s²

g = Acceleration due to gravity at earth's surface = 9.8 m/s²

R = Radius of earth = 6400 km

h = Height

On substituting the values, we get,

6/9.8 = (6400)²/(6400 + h)²

0.612 = (40.96 × 10⁶)/((40.96 × 10⁶) + h² + (12.8 × 10³)h)

0.612 × ((40.96 × 10⁶) + h² + (12.8 × 10³)h) = 40.96 × 10⁶

(25.06 × 10⁶) + 0.612h² + (7.83 × 10³)h = 40.96 × 10⁶

0.612h² + (7.83 × 10³)h = (40.96 × 10⁶) - (25.06 × 10⁶)

0.612h² + (7.83 × 10³)h = 15.9 × 10⁶

0.612h² + (7.83 × 10³)h - (15.9 × 10⁶) = 0

On applying quadratic equation given below, we get,

h = (-b ± √(b² - 4ac))/2a

Where,

a = 0.612; b = 7.83 × 10³; c = -(15.9 × 10⁶)

On applying quadratic equation, we get,

h = (-0.612 ± √((0.612)² + (4 × 0.612 × 15.9 × 10⁶)))/(2 × 0.612)

On solving, we get,

h = 1782.35 or −14576.5

The negative acceleration due to gravity is impossible. Thus, h is 1782.35 km.