Question

Different satellites orbit the earth with a vast range of altitudes, from just a couple hundred km, all the way to tens of thousands of km above the surface. The international space station (ISS) is in a low earth orbit, just 400km above the surface (you can see it with the naked eye at sunset and sunrise as a bright, moving dot). At this altitude, the acceleration due to gravity has a value of 8.69.

A. What is the speed of the ISS?
B. What is the orbital period (T) of the ISS in minutes?

Answer 1

The speed of the ISS is 7690 km / s

The orbital period (T) of the ISS in minutes is 93 minutes

How to solve for the speed

The speed of the ISS is given as

`√(g) *(r + h)`

In the equation we

r + h = 6400 + 400

= 6800

such that

√8.96 * 6800x10³

= 7690 km / s

The orbital period

= 2 x π x r / v

= 2 x π x 6800 / 7690

= 5.55 x 10 ³ seconds

= 93 minutes

The orbital period (T) of the ISS in minutes is 93 minutes

Answer 2

Find the answer in the explanation.

Step-by-step explanation:

Given that the international space station (ISS) is in a low earth orbit, just 400km above the surface. And at this altitude, the acceleration due to gravity has a value of 8.69.

A.) Since the radius of the earth is equal to 6400 km. At it to 400km

The radius of the orbit = 6400 + 400

The radius = 6800 km.

Where orbital speed = sqrt ( g × r )

Orbital speed = sqrt ( 6800 × 8.69 )

Orbital speed = 243.09 km/h

B.) Orbital period acceleration to Kepler third law of planetary motion state that:

The square of the period is directly proportional to the cube of the radius.

T^2 = (4π^2 /GM) r^3

T^2=(4π^2/ 6.67×10^11 × 6.0 × 10^24)r^3

T^2 = (4π^2/ 4.002^36) × 6800^3

T^2 = 9.865×10^-36 × 6800^3

T^2 = 3.10 × 10^-15

T = 5.57 × 10^-8 hours.

Convert the hours to minutes by multiplying it by 60

T = 3.4 × 10^-6 minutes.