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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?
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?

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0we 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what would be my Re and Me?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it 8.42 x10 ^23?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it 3.021 x 1034?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0i 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.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0from 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 \(\frac{R_x}{R_e}\)
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