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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?

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  1. IrishBoy123
    • one year ago
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    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
    • one year ago
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    what would be my Re and Me?

  3. anonymous
    • one year ago
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    is it 8.42 x10 ^-23?

  4. anonymous
    • one year ago
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    @IrishBoy123

  5. anonymous
    • one year ago
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    is it 3.021 x 10-34?

  6. IrishBoy123
    • one year ago
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    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
    • one year ago
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    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 \(\frac{R_x}{R_e}\)

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