Question

If the earth were a sphere made completely of gold, what would have been the magnitude of gravitational acceleration on its surface? The radius of the earth = 6400 km, density of gold = 19.3 × 10³ kg/m³, G = 6.67 × 10⁻¹¹ Nm²/kg².

Answer 1

Answer:

3.5ms^-2

Explanation:

Refer to the material.

Answer 2

Answer:

$g=   3.5 m/sec^{2}$

Explanation:

Here, in the given question

Radius of the earth is = 6400 Km

Density of gold = 19.3× $10^{-3} kg/m^{3}$

Nd G = 6.67×$10^{-11}Nm^{2}/kg^{2}$

Density = $\dfrac{mass}{Volume}$

we have the formula for gravitational acceleration

$mg= \dfrac{GMm}{R^{2} } \\So,\\g= \dfrac{GM}{R^{2} } \\And,\\Density =\dfrac{mass}{volume}\\ Mass (M) = Density * volume$

Putting the all the given values in the formula

$g= \dfrac{G*density*volume}{R^{2} } \\And, volume= \frac{4}{3}R^{3}$π

$g=\dfrac{6.67*10^{-11}*19.3* 10^{-3}*4*3.14*6400*6400*6400}{6400*6400*3} \\g=   3.5 m/sec^{2}$