Question

A hypothetical Planet X of mass 7.5 ✕ 1023 kg rotates on its axis once every 5 days. Hint (a) Find the distance from the center of Planet X (in m) at which a satellite would remain in one spot above the planet's surface. m (b) Find the speed of the satellite (in m/s). m/s

Answer 1

Answer:

(A) The distance of the center of Planet X is $6.18 \times 10^{7}$ m

(B) The speed of the satellite is $898.38 \frac{m}{s}$

Explanation:

Given:

Mass of the planet $m = 7.5 \times 10^{23}$ kg

Time period $T = 5$ days $= 432000$ sec

(A)

From kepler's third law

 $T^{2} = \frac{4\pi ^{2}r^{3} }{Gm}$

Where $G = 6.67 \times 10^{-11}$

Now for finding the distance of planet X,

  $r^{3} = \frac{Gm T^{2} }{4\pi ^{2} }$

  $r^{3} = \frac{6.67 \times 10^{-11} \times 7.5 \times 10^{23 } \times (432000) ^{2} }{4 (3.14)^{2} }$

  $r^{3} = 236 \times 10 ^{21}$

   $r = (236 \times 10^{21} )^{\frac{1}{3} }$

   $r = 6.18 \times 10^{7}$ m

(B)

Speed of the satellite is given by,

   $v = \frac{2\pi r}{T}$

   $v = \frac{6.28 \times 6.18 \times 10^{7} }{432000}$

   $v = 898.38$ $\frac{m}{s}$

Therefore, the speed of planet is $898.38 \frac{m}{s}$