Question

A newly discovered planet has a mass 1.5 times compared to that of the earth, and its diameter is 3 times compared to earth. What is the gravitational acceleration on this planet?

Answer 1

we relate the idea of g at the earth's surface to Newton's law of universal gravitation, such that at the surface of our planet of radius \(R_e\), we can say:

\(\large F = \frac{GM_em}{R_e^2} = mg_e\)

so

\(\large g_e = \frac{GM_e}{R_e^2}\)

Then, noting that G is a universal constant, we can say that:

\(\large G = \frac{g_e \ R_e^2}{M_e}\)



such that, as between 2 different planets, earth and planet X:

\( \huge \frac{g_e \ R_e^2}{M_e} = \frac{g_x \ R_x^2}{M_x} \)


and so, to find \(g_x\) using the data provided in the question

Answer 2

what would be my Re and Me?

Answer 3

is it 8.42 x10 ^-23?

Answer 4

@IrishBoy123

Answer 5

is it 3.021 x 10-34?

Answer 6

i was teasing. my apologies.

we know from above equations that:

\(\large g_x = g_e \times \frac{R_e^2}{Me} \times \frac{M_x}{R_x^2} = g_e \times (\frac{R_e}{R_x})^2 \times \frac{M_x}{M_e}\)

you have everything [ie the ratios!!] you need in the original question to do this.

Answer 7

from the question:

.....**has a mass 1.5 times compared to that of the earth***..... ***diameter is 3 times compared to earth****

so you know \(\frac{M_x}{M_e}\) and \(\frac{R_x}{R_e}\)