If the planet Mercury has a mass of planet 3.3×10²³ kg and a radius of 2400 km - calculate the magnitude of the gravitational field on its surface?

Respuesta :

Answer:

The magnitude of the gravitational field on the surface of the planet mercury is 3.82 m/s²

Explanation:

The mass of mercury, m = 3.3×10²³ kg

The radius, r = 2400 km

r = 2400 x 1000m

r = 2.4 x 10⁶m

Note that the magnitude of the gravitational field on the surface of the planet is the acceleration due to gravity on that planet

It is given by the formula:

[tex]g=\frac{Gm}{r^2}[/tex]

Substitute the values of G, m, and r into the formula above

[tex]\begin{gathered} g=\frac{6.67\times10^{-11}\times3.3\times10^{23}}{(2.4\times10^6)^2} \\ g=\frac{6.67\times10^{-11}\times3.3\times10^{23}}{(2.4\times10^6)^2} \\ g=\frac{2.2\times10^{13}}{5.76\times10^{12}} \\ g=3.82m/s^2 \end{gathered}[/tex]

The magnitude of the gravitational field on the surface of the planet mercury is 3.82 m/s²