Question

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?

Answer 1

Answer:

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

Step-by-step 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:

`g=(Gm)/(r^2)`

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

`\begin{gathered} g=(6.67*10^(-11)*3.3*10^(23))/((2.4*10^6)^2) \\ g=(6.67*10^(-11)*3.3*10^(23))/((2.4*10^6)^2) \\ g=(2.2*10^(13))/(5.76*10^(12)) \\ g=3.82m/s^2 \end{gathered}`

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