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 $898.38 \frac{m}{s}$

Answer 2

The motion of the satellite around the hypothetical Planet X is described

by Kepler's laws of planetary motion.

(a) The distance from the center of the center of Planet X at which the

satellite would remain in one spot is approximately 61852641 meters.

(b) The speed of the satellite is approximately 899.6 m/s

Reasons:

The given parameters are;

Mass of Planet X, M = 7.5 × 10²³ kg.

Number of times the plane rotates every 5 days = Once

(a) The period of the satellite around a planet is given by the formula;

According to Kepler's third law of planetary motion, we have;

$T^2 \propto r^3$

$T^2 = \left(\dfrac{4 \cdot \pi ^2}{G \cdot M} \right) r^3$

Therefore;

$r = \sqrt[3]{\dfrac{T^2 \cdot G \cdot M}{4 \cdot \pi^2} }$

Where;

T = 5 days = 432,000 seconds

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

Which gives;

$r = \sqrt[3]{\dfrac{432000^2 \times 6.67430 \times 10^{-11} \times 7.5 \times 10^{23}}{4 \cdot \pi^2} } \approx 61852641$

The distance from the center of the center of Planet X at which the satellite

would remain in one spot, r ≈ 61852641 meters

(b) The angular speed of the satellite, ω, is given as follows;

$\omega = \dfrac{2 \cdot \pi}{432000} \ rad/s$

The linear speed, v = ω × r

Therefore;

$v = \omega = \dfrac{2 \cdot \pi}{432000} \times 61852641 \approx 286.35 \cdot \pi \approx 899.6$

The speed of the satellite, v ≈ 899.6 m/s

Learn more here:

https://brainly.com/question/22424027