HELP ME WITH THIS FOR A BRAINLIEST

The international Space Station (ISS) orbits the Earth once every 90 mins at an altitude of 409 km. How high would it have to be

Oksi-84 [34.3K]

It would have to be 36,719 Km high in order to be to be in geosynchronous orbit.

To find the answer, we need to know about the third law of Kepler.

<h3>What's the Kepler's third law?</h3>

It states that the square of the time period of orbiting planet or satellite is directly proportional to the cube of the radius of the orbit.
Mathematically, T²∝a³

<h3>What's the radius of geosynchronous orbit, if the time period and altitude of ISS are 90 minutes and 409 km respectively?</h3>

The time period of geosynchronous orbit is 24 hours or 1440 minutes.
As the Earth's radius is 6371 Km, so radius of the ISS orbit= 6371km + 409 km = 6780km.
If T1 and T2 are time period of geosynchronous orbit and ISS orbit respectively, a1 and a2 are radius of geosynchronous orbit and ISS orbit, as per third law of Kepler, (T1/T2)² = (a1/a2)³
a1= (T1/T2)⅔×a2

= (1440/90)⅔×6780

= 43,090 km

Altitude of geosynchronous orbit = 43,090 - 6371= 36,719 km

Thus, we can conclude that the altitude of geosynchronous orbit is 36,719km.

Learn more about the Kepler's third law here:

brainly.com/question/16705471

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6 0

2 years ago

The net force of a heavy, stationary box is?

Leto [7]

Answer:

gravity

Explanation:

7 0

3 years ago

A toy helium balloon is initially at a temperature of T = 24o C. Its initial volume is 0.0042 m3 (It is a 10 cm radius sphere).

Anettt [7]

Answer:

$T_{f} = 335.780\,K\,(62.630\,^{\textdegree}C)$

Explanation:

Let assume that air behaves ideally. The equation of state of ideal gases is:

$P\cdot V = n\cdot R_{u}\cdot T$

Where:

$P$ - Pressure, in kPa.

$V$ - Volume, in m³.

$n$ - Quantity of moles, in kmol.

$R_{u}$ - Ideal gas constant, in $\frac{kPa\cdot m^{3}}{kmol\cdot K}$ .

$T$ - Temperature, in K.

Since there is no changes in pressure or the quantity of moles, the following relationship between initial and final volumes and temperatures is built:

$\frac{V_{o}}{T_{o}} = \frac{V_{f}}{T_{f}}$