Given: G = 6.67259 × 10−11 N m2 /kg2 . A 438 kg geosynchronous satellite orbits a planet similar to Earth at a radius 1.94 × 105 km from the planet’s center. Its angular speed at this radius is the same as the rotational speed of the Earth, and so they appear stationary in the sky. That is, the period of the satellite is 24 h . What is the force acting on this satellite? Answer in units of N.

[tex]T=2\pi\sqrt{\frac{r^3}{GM}}\\\Rightarrow M=4\pi^2\frac{r^3}{T^2G}\\\Rightarrow M=4\pi^2\times \frac{(1.94\times 10^8)^3}{(24\times 3600)^2\times 6.67259\times 10^{-11}}\\\Rightarrow M=5.78686\times 10^{26}\ kg[/tex]

The gravitational force is given by

[tex]F=G\frac{Mm}{r^2}\\\Rightarrow F=6.67259\times 10^{-11}\times \frac{5.78686\times 10^{26}\times 438}{(1.94\times 10^8)^2}\\\Rightarrow F=449.37412\ N[/tex]

The force acting on this satellite is 449.37412 N

ap8997154 ap8997154 · Answer 2 · 2022-03-16T10:24:45+01:00

Gravity, is often known as gravitation. The gravitational force between the planet and the satellite can be written as 0.4332 N.

What is gravitational force?

Gravity, often known as gravitation, is the universal force of attraction that acts between all matter in mechanics. It is the weakest known force in nature, and so has no bearing on the interior properties of ordinary matter.

[tex]F =G\dfrac{m_1m_2}{r^2}[/tex]

As it is given that the value of the gravitational constant is G = 6.67259 × 10−11 N m² /kg², while the radius of the planet is 1.94×10⁵. And the time taken by the satellite to revolve around the planet is 24 hours. therefore, the Mass of the planet can be written as,

The Time period of the satellite is given as:

[tex]T = 2\pi \sqrt{\dfrac{r^3}{GM}}[/tex]

Substitute the values in the formula,

[tex]24 = 2\pi \sqrt{\dfrac{(1.94 \times 10^5)^3}{6.67259 \times 10^{-11}M}}\\\\M = 5.78686 \times 10^{17}\rm\ kg[/tex]

Thus, the mass of the planet can be written as 5.5786×10¹⁷ kg.

Now, the gravitational force can be written as,

[tex]F = G\dfrac{Mm}{r^2}\\\\F = 6.67259\times 10^{-11} \times \dfrac{5.5786 \times 10^{17} \times 438}{1.94 \times 10^5}\\\\ F = 0.4332\rm\ N[/tex]

Hence, the gravitational force between the planet and the satellite can be written as 0.4332 N.

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