Assume that the Moon is at a distance of3,00,000 km from the Earth, and that ittakes 28 days for it to orbit the Earth once.Geostationary satellites are those which areat a rest relative to the Earth. Using thesetwo facts, derive the altitude of thegeostationary satellite from the centre ofthe Earth.
Question
Assume that the Moon is at a distance of3,00,000 km from the Earth, and that ittakes 28 days for it to orbit the Earth once.Geostationary satellites are those which areat a rest relative to the Earth. Using thesetwo facts, derive the altitude of thegeostationary satellite from the centre ofthe Earth.
Solution
To derive the altitude of a geostationary satellite from the center of the Earth, we need to consider the distance between the Moon and the Earth and the orbital period of the Moon.
Given that the Moon is at a distance of 3,00,000 km from the Earth and it takes 28 days for it to orbit the Earth once, we can use this information to calculate the altitude of the geostationary satellite.
First, we need to convert the orbital period of the Moon from days to seconds. There are 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute. So, there are 24 * 60 * 60 = 86,400 seconds in a day.
Therefore, the orbital period of the Moon in seconds is 28 * 86,400 = 2,419,200 seconds.
Next, we can use the formula for the orbital period of a satellite to calculate the altitude. The formula is:
T = 2π√(r^3/GM)
Where T is the orbital period, r is the distance from the center of the Earth to the satellite, G is the gravitational constant, and M is the mass of the Earth.
Since we want the satellite to be at rest relative to the Earth, we can set the orbital period T equal to 24 hours, which is the time it takes for the Earth to complete one rotation.
Plugging in the values, we have:
24 * 60 * 60 = 2π√(r^3/GM)
Simplifying the equation, we get:
r^3 = (24 * 60 * 60)^2 * (GM) / (4π^2)
Now, we can substitute the known values. The mass of the Earth (M) is approximately 5.972 × 10^24 kg, and the gravitational constant (G) is approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2.
Plugging in these values, we have:
r^3 = (24 * 60 * 60)^2 * (6.67430 × 10^-11 * 5.972 × 10^24) / (4π^2)
Simplifying further, we get:
r^3 = 4.203 × 10^14
Taking the cube root of both sides, we find:
r ≈ 3,981 km
Therefore, the altitude of the geostationary satellite from the center of the Earth is approximately 3,981 km.
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