# On July 15, 2004, NASA launched the Aura spacecraft to study the earth's climate and atmosphere....

## Question:

On July 15, 2004, NASA launched the Aura spacecraft to study the earth's climate and atmosphere. This satellite was injected into an orbit 705 km above the earth's surface, and we shall assume a circular orbit. How many hours does it take this satellite to make one orbit? How fast (in km/s) is the Aura spacecraft moving?

## Satellite Speed and Time Period:

{eq}\\ {/eq}

A satellite is any object that orbits around a celestial body solely due to the gravitational force of the celestial body on the satellite. The gravitational force on the satellite, which depends on the mass of the celestial body, the mass of the satellite, and the distance between them, acts as a centripetal force for the circular motion of the satellite.

The speed of the object in a uniform circular motion is directly related to the centripetal force acting on the object, the radius of the circle, and the mass of the object. Hence the speed of a satellite is directly related to the factors on which the centripetal force on the satellite depends.

The period of the satellite, which is equal to the time taken by the satellite to complete one orbit also depends on the above-mentioned factors.

{eq}\\ {/eq}

We are given:

• The altitude of the satellite from the surface of Earth, {eq}h=705\;\rm km=7.05\times 10^{5}\;\rm m {/eq}

The time required to complete an orbit is called the period of the satellite.

The time period of the motion of a satellite around a planet depends only on the mass, {eq}M {/eq}, of the planet and its distance, {eq}r {/eq}, from the center of the planet by the following equation:

{eq}T=2\pi\sqrt {\dfrac{r^3}{GM}} {/eq}

Here,

• {eq}G=6.67\times 10^{-11}\;\rm m^3/kg.s^2 {/eq} is the gravitational constant.

The mass of Earth is: {eq}M=5.98\times 10^{24}\;\rm kg {/eq}

The radius of the Earth is {eq}R=6.37\times 10^{3}\;\rm km=6.37\times 10^{6}\;\rm m {/eq}

Therefore, the distance of the satellite from the center of Earth is:

{eq}\begin{align*} r&=R+h\\ &=\left ( 6.37+0.705 \right )\times 10^{6}\;\rm m\\ &\approx 7.08\times 10^{6}\;\rm m \end{align*} {/eq}

Therefore, the period of the satellite is:

{eq}\begin{align*} T&=2\pi\sqrt {\dfrac{\left ( 7.08\times 10^{6} \right )^3}{6.67\times 10^{-11}\times 5.98\times 10^{24}}}\\ &=2\pi\sqrt {88.976\times 10^{4}}\\ &=2\pi\times 9.43\times 10^{2}\;\rm s\\ &=\boxed{5.92\times 10^{3}\;\rm s}\\ &=\dfrac{5.92\times 10^{3}}{3.6\times 10^{3}}\;\rm h\\ &=\boxed{1.64\;\rm h} \end{align*} {/eq}

The speed of a satellite depends only on the mass of the planet/star, {eq}M {/eq}, orbited by the satellite, and the radius, {eq}r {/eq}, of the circular orbit followed by the satellite by the following equation:

{eq}v=\sqrt{\dfrac{GM}{r}} {/eq}

Plugging in the given values, we have:

{eq}\begin{align*} v&=\sqrt{\dfrac{6.67\times 10^{-11}\times 5.98\times 10^{24}}{7.05\times 10^{6}}}\\ &=\sqrt{56.58\times 10^{6}}\\ &=\boxed{7.52\times 10^{3}\;\rm m/s} \end{align*} {/eq}