The length of the day on planet x would be 2.65 * 10⁴. On the other hand, the orbital speed of the satellite would be: 8.9 * 10³.
To find the length of the day on planet X we must take into account the information provided in the question and carry out the following mathematical procedure:
The distance between North Pole and equator over the surface that is equal to the one-fourth of the circumference.
[tex]d = \frac{2\pi R}{4} \\\\R= \frac{2d}{\pi} \\= \frac{(18850 km)}{\pi } \\\\= 12000 km[/tex]
Therefore,
[tex]8.837 m/s^{2} = 9.509 m/s^{2} - (12000 * 10^{3} m ) 2 cos (rad)\\\\\\\sqrt{\frac{9.509 m/s^{2} - 8.8357 m/s^{2} }{12000 * 10^{3 m} } } \\\\= 2.2368 * 10^{-4} rad/s[/tex]
Therefore, the time period of the planet is
[tex]T = \frac{2\pi }{2.368 * 10^{-4} rad/s} \\= 2.65 * 10^{4} seg[/tex]
This is the time of one day in the planet X.
To calculate the orbital speed of the satellite we must take into account the data provided by the question and carry out the following procedure:
The mass of the planet X is,
M = gx-northpole R 2/G
[tex]M = \frac{gx - north pole R2}{ G}\\= \frac{(9.509 m/s^{2})(12000 * 10^{3})}{6.67 * 10 x^{-11} N * m^{2}/kg^{2} } \\= 2.053 * 10^{25} kg\\[/tex]
The orbital speed of the satellite is,
[tex]Vo = \sqrt[2]{x} \frac{GM}{r} \\= \sqrt[n]{\frac{(6..67*10^{-11})(2.053 * 1025kg}{12000*10^{3} + 2000 * 10^{3}) m } } \\= 9.88 * 10 x^{3} m/s[/tex]
The orbital speed of the satellite is,
[tex]T = \frac{2\pi r}{Vo} \\= \frac{2\pi (12000*10^{3} m + 2000 * 10^{3} m }{9.88 * 10^{3} m/s } \\=8.9 * 10^{3} s[/tex]
Learn more about planet X in: https://brainly.com/question/14581221
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