Question

Calculate the effective value of g, the acceleration of gravity, at 6700 m , above the Earth's surface. g

Answer 1

Answer:

The effective value of g at 6700 m above the Earth's surface is 9.79 m/s².

Explanation:

The value of g can be found using the following equation:

$F = \frac{GmM}{r^{2}}$

$ma = \frac{GmM}{r^{2}}$

$a = \frac{GM}{r^{2}}$

Where:

a is the acceleration of gravity = g

G: is the gravitational constant = 6.67x10⁻¹¹ m³/(kg.s²)

M: is the Earth's mass = 5.97x10²⁴ kg

r: is the Earth's radius = 6371 km      

Since we need to find g at 6700 m, the total distance is:

$r_{T} = 6371000 m + 6700 m = 6377700 m$

Now, the value of g is:

$a = \frac{GM}{r_{T}^{2}} = \frac{6.67\cdot 10^{-11} m^{3}/(kg*s^{2})*5.97 \cdot 10^{24} kg}{(6377700 m)^{2}} = 9.79 m/s^{2}$

Therefore, the effective value of g at 6700 m above the Earth's surface is 9.79 m/s².

I hope it helps you!