Question

calculate the stationary satellite orbital period that
revolves around the Earth in
circular orbit and
focus on the Earth's Position the gravity accleration
is 8.50 m/s^2

1 ) 1.20 h
2) 1.57 h
3) 2.35 h
4) 3.36 h
please ans​

Answer 1

Answer:

1.20h

Explanation:

it is due to gravity

Answer 2

Answer:

Explanation:

Concept:

The concept of gravity will be used to solve this question.The force that pulls items toward the centre of a planet or other entity is called gravity. The gravitational pull of the sun keeps every planet in its orbit around it.

Given:

The gravity acceleration is $8.50 m/s^{2}$ .

To Find:

We need to find the the stationary satellite orbital period that revolves around the Earth in circular orbit .

Solution:

The altitude at which the gravity has value $8.50 m/s^{2}$ is,

$g_{h}=\frac{g}{\left(1+\frac{h}{R}\right)^{2}}$

First we will calculate  the denominator.

Put the value in the denominator.So we can write

$\left(1+\frac{h}{R}\right)=\sqrt{\frac{9.80 \mathrm{~m} / \mathrm{s}^{2}}{8.50 \mathrm{~m} / \mathrm{s}^{2}}}$

By solving we will get $1+\frac{h}{R}=1.07375$

Take $1$ to the right side and calculate the value of $R$ .

$\frac{h}{R}=0.07375$

The value of $R$ is $6.37 \times 10^{6} \mathrm{~m}$ .

So the value of $h$ is $h= 0.07375 \times 6.37 \times 10^{6} \mathrm{~m} \approx 4.70 \times 10^{5} \mathrm{~m}$ .

The formula of the time period is $T=2 \pi \sqrt{\frac{r^{3}}{G M_{E}}}$

$r^{3}$ can be written as $(R+h)^{3}$.

So the formula is $T=2 \pi \sqrt{\frac{(R+h)^{3}}{G M_{E}}}$

Put the value in the above equation.

$T=2 \pi \sqrt{\frac{\left(6.37 \times 10^{6} \mathrm{~m}+4.70 \times 10^{5} \mathrm{~m}\right)^{3}}{\left(6.67 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2}\right)\left(5.972 \times 10^{24} \mathrm{~kg}\right)}}$,

Solve equation to get the value of $T$ is $T=5631.7 \mathrm{~s}\left(\frac{1 \mathrm{hr}}{3600 \mathrm{~s}}\right)$

The value of $T$ is $1.56 \mathrm{hr}$ .

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