A satellite circles the earth in an orbit whose radius is 2.07 times the earth's radius. The earth's mass is 5.98 x 1024 kg, and its radius is 6.38 x 106 m. What is the period of the satellite?

Well, if this satellite circles the Earth, it must be a really dedicated stalker. I hope it has its own space restraining order! Anyway, to calculate the period of the satellite, we can use Kepler's Third Law, which states that the square of the period of a satellite is proportional to the cube of the semi-major axis of its orbit. In this case, the semi-major axis of the orbit is 2.07 times the Earth's radius. So, let's calculate it. 2.07 times 6.38 x 10^6 meters equals... wow, that's a big number! Anyway, once we figure out the semi-major axis, we can use it to calculate the period of the satellite using Kepler's Third Law. So, do some math and you'll have your answer! Just remember, the satellite might enjoy the ride, but it might also need some space from Earth every now and then.

To find the period of the satellite, we can use Kepler's third law of planetary motion, which states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

Given:
Radius of the satellite's orbit (r) = 2.07 * radius of the earth (R) = 2.07 * 6.38 x 10^6 m = 1.31866 x 10^7 m.
Mass of the earth (M) = 5.98 x 10^24 kg.
Radius of the earth (R) = 6.38 x 10^6 m.

Using the equation for the period of the satellite:
T^2 = (4π^2 / GM) * r^3,

Where:
T is the period of the satellite.
G is the gravitational constant (6.67430 x 10^-11 m^3 kg^-1 s^-2).
M is the mass of the earth.
r is the radius of the satellite's orbit.

Plugging in the values:
T^2 = (4π^2 / (6.67430 x 10^-11 * 5.98 x 10^24)) * (1.31866 x 10^7)^3,

Simplifying the equation:
T^2 ≈ 4.0 x 10^-20 * (1.31866 x 10^7)^3,

Calculating:
T^2 ≈ 4.0 x 10^-20 * 2.61 x 10^21,
T^2 ≈ 1.044 x 10^2.

Taking the square root of both sides gives us:
T ≈ sqrt(1.044 x 10^2),
T ≈ 10.22 seconds.

Therefore, the period of the satellite is approximately 10.22 seconds.

To find the period of the satellite, we can use the formula:

\[T = 2\pi \sqrt{\frac{r^3}{G M}}\]

Where:
T = Period of the satellite
r = Radius of the satellite's orbit
G = Gravitational constant (approximately 6.674 × 10^-11 m^3 kg^-1 s^-2)
M = Mass of Earth

First, we need to calculate the radius of the satellite's orbit, which is given as 2.07 times the radius of the Earth.

Given:
Radius of the Earth (R) = 6.38 x 10^6 m
Radius of the satellite's orbit (r) = 2.07 * Radius of Earth

To find the radius of the satellite's orbit:
\(r = 2.07 \times 6.38 \times 10^6 \, \text{m}\)

Next, we can substitute the values into the formula to find the period of the satellite:

\[T = 2\pi \sqrt{\frac{(2.07 \times 6.38 \times 10^6)^3}{6.674 \times 10^{-11} \times 5.98 \times 10^{24}}} \]

Now, let's calculate the value to find the period of the satellite.

the gravitational constant G =6.67•10⁻¹¹ N•m²/kg²,

Earth’s mass is M = 5.98•10²⁴kg,
Earth’s radius is R = 6.38•10⁶ m.

T=L/v

F = m₁•m₂/R= G•m•M/(2.07R) ²
F=ma=m•v²/(2.07R)
G•m•M/(2.07R)² =m•v²/(2.07R)
v=sqrt{G•M/2.07R}

T=2π(2.07R)/sqrt{G•M/2.07R}=…