91Ó°ÊÓ

The given data can be listed below as

• Weight of astronaut on the earth$w_e=943.0\mathrm{N}$
• Weight of astronaut on the north of planet X,$w_{xn}=915.0\mathrm{N}$
• Weight of astronaut on the equator of planet X,$w_{xe}=850.0\mathrm{N}$
• Distance or circumference from the north pole to the equator of planet$Xd/4=18,850\mathrm{Km}$
• Weight of satellite$m_s=45,000\mathrm{Kg}$

· Distance of the satellite from the planet$X,a=2000\mathrm{Km}$

Kepler has given three laws for the orbital movement of celestial bodies around any heavenly body in space. The 3 law that is related to the period of the orbital movement states that,

The period or period of the movement of any celestial body around another heavenly body is directly proportional to the three-by-two power of the radius or distance between the objects.

To find out the radius r of planet X, we will apply the formula of the circumference d of the sphere, that is

$$\begin{gathered} d=2\mathrm{τ}\mathrm{τ}r \\ r=\frac{d}{2\mathrm{τ}\mathrm{τ}} \\ r=\frac{18850×4×7}{2×22}\mathrm{Km} \\ r=11995.45\mathrm{Km} \end{gathered}$$

To find out the mass of the planet X, we will use newton’s law of gravitational force, that is

$$\begin{gathered} g=\frac{G{m}_x}{r^2} \\ {m}_x=\frac{g{r}^2}{G} \\ {m}_x=\frac{\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)×{\left(11995450\mathrm{m}\right)}^2}{\left(6.67×{10}^{-11}{\mathrm{m}}^3{\mathrm{kg}}^{-1}{\mathrm{s}}^{-2}\right)} \\ {m}_x=2.1×{10}^{25}\mathrm{kg} \end{gathered}$$

Here, g is gravity, G is gravitational constant, r is the radius, $m_x$is the mass of planet X.

From the Kepler 3rd law, the period

localid="1657180093694" $$\begin{gathered} T=\frac{2\mathrm{τ}\mathrm{τ}{r}^{3/2}}{\sqrt{G{m}_x}} \\ T=\frac{{\displaystyle 2×22/7}×{\left(11995450\mathrm{m}\right)}^{3/2}}{\sqrt{6.67×{10}^{-11}{\mathrm{m}}^3{\mathrm{Kg}}^{-1}{\mathrm{s}}^{-2}×2.1×{10}^{25}\mathrm{kg}}} \\ T=1864\sec \end{gathered}$$

Here, Tis the period, r is the radius of planet X, G is the gravitational constant, and $m_x$is the mass of planet X.

Hence, the length of a day on planet X is 1864 sec.

From Kepler’s 3rd law, the orbital period of the satellite,

$$\begin{gathered} T=\frac{2\mathrm{τ}\mathrm{τ}{a}^{3/2}}{\sqrt{G\left(m_x+m_s\right)}} \\ T=\frac{{\displaystyle 2×22/7×{\left(2000000\mathrm{m}\right)}^{3/2}}}{\sqrt{6.67×{10}^{-11}{\mathrm{m}}^3{\mathrm{Kg}}^{-1}{\mathrm{s}}^{-2}×\left(2.1×{10}^{25}+45000\right)\mathrm{kg}}} \\ T=126\sec \end{gathered}$$

Here,Tis the period, r is the radius of planet X, G is the gravitational constant, and $m_x$is the mass of Planet X$m_s$is the Weight of the satellite.

Hence, the orbital period of the satellite is 126 sec.