Physics: A round asteroid has a surface gravity of .0288 m/s^2 if the mass of the asteroid is 1.10 x10^18 kg what is its radius

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A round asteroid has a surface gravity of .0288

Ответ:

jscout2468

12.06.2021

Physics

r = 50.47 x 10³ m = 50.47 km

Explanation:

Using the formula for the acceleration due to gravity:

$g = \frac{Gm}{r^2}$

where,

g = acceleration due to gravity on the surface of asteroid = 0.0288 m/s²

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ Nm²/kg²

m= mass of asteroid = 1.1 x 10¹⁸ kg

r = radius of asteroid = ?

Therefore,

$0.0288\ m/s^2 = \frac{(6.67\ x\ 10^{-11}\ Nm^2/kg^2)(1.1\ x\ 10^{18}\ kg)}{r^2} \\\\r = \sqrt{\frac{(6.67\ x\ 10^{-11}\ Nm^2/kg^2)(1.1\ x\ 10^{18}\ kg)}{0.0288\ m/s^2}}$

r = 50.47 x 10³ m = 50.47 km

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A round asteroid has a surface gravity of .0288 m/s^2 if the mass of the asteroid is 1.10 x10^18 kg what is its radius

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