Question

A satellite revolves around the earth in a circular orbit of radius r=4×1016 m. the period of the satellite is:

Answer 1

1,680 days

Step-by-step explanation:

We are given that the satellite revolves in a circular orbit with radius r = 4 x 1016 meters. The unit of 1016 means the radius is 4 trillion meters.

We want to find the period or time taken for one revolution. We can use Kepler's third law to calculate the period.

Kepler's third law relates the period (P) of an orbiting body to the semi-major axis (a) of its orbit:

• P2 ∝ a3

Where a is equal to the radius r for a circular orbit.

The proportionality constant depends on the large scale gravitational parameter (μ) of the central body (in this case, Earth). The law can be written as:

P2 = (4π2/μ) * a3

Where μ for Earth is 398,600 km3/s2. Converting the radius r in meters to kilometers:

a = 4 x 1016 m = 4 x 109 km

Substituting the values of a and μ in the law, we get:

P2 = (4π2/398600)*(43)*109)^3

P2 = 1.62 x 1024

Taking the square root, we get the period:

P = 1.27 x 1012 seconds

Converting this to hours gives the final answer:

1.27 x 1012 seconds / (60*60 seconds/hour) =

1.48 x 105 hours

Therefore, the period (time for one revolution) of the satellite is around 1.48 x 10^5 hours or 1,680 days.

Answer 2

The period of a satellite in a circular orbit around the Earth can be calculated using the formula:

T = 2π√(r³/GM)

where:

T = period of the satellite

r = radius of the orbit

G = gravitational constant (6.67 x 10^-11 Nm^2/kg^2)

M = mass of Earth (5.97 x 10^24 kg)

Substituting the given values:

T = 2π√((4x10^16)^3 / (6.67x10^-11 x 5.97x10^24))

T = 2π√(135168 x 10^22 / 3.98739 x 10^14)

T = 2π√(338828.41)

T = 2 x 3.1416 x 582.3408

T = 3661.6 seconds

Therefore, the period of the satellite is 3661.6 seconds, or approximately 61 minutes.