## anonymous one year ago 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?

1. IrishBoy123

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

2. anonymous

what would be my Re and Me?

3. anonymous

is it 8.42 x10 ^-23?

4. anonymous

@IrishBoy123

5. anonymous

is it 3.021 x 10-34?

6. IrishBoy123

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.

7. IrishBoy123

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 $$\frac{R_x}{R_e}$$