# A satellite is placed in orbit 5.20 x 10^5 m above the surface of the planet Jupiter. Jupiter has...

## Question:

A satellite is placed in orbit 5.20 x {eq}10^5 {/eq} m above the surface of the planet Jupiter. Jupiter has a mass of 1.90 x {eq}10^{27} {/eq} kg and a radius of 7.14 x {eq}10^7 {/eq} m. Find the orbital speed of the satellite.

## Speed of a Satellite:

{eq}\\ {/eq}

A satellite is a body that orbits a planet/moon/star. The centripetal force needed for this circular motion is provided by the gravitational force exerted by the planet.

The centripetal force experienced by an object depends on its mass, speed, and the radius of the motion; and the gravitational force between two objects also depends on the masses of the individual objects, and the distance between them. Because of this dependence, it turns out that the speed of the satellite in an orbit depends only on the radius of the orbit, and the mass of the orbiting planet.

{eq}\\ {/eq}

We are given:

• The mass of the Jupiter, {eq}M=1.90\times 10^{21}\;\rm kg {/eq}
• The radius of the Jupiter, {eq}R=7.14\times 10^{7}\;\rm m {/eq}
• The height of the satellite from the surface of Jupiter, {eq}h=5.2\times 10^{5}\;\rm m=0.052\times 10^{7}\;\rm m {/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}

Here,

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

The radius of the orbit of the satellite is:

{eq}\begin{align*} r&=R+h\\ &=0.052\times 10^{7}+7.14\times 10^{7}\\ &=7.192\times 10^{7}\;\rm m \end{align*} {/eq}

After plugging the given values into the above equation, we have:

{eq}\begin{align*} v&=\sqrt{\dfrac{6.67\times 10^{-11}\times 1.90\times 10^{27}}{7.192\times 10^{7}}}\\ &=\sqrt{17.62\times 10^{8}}\\ &\approx \boxed{4.2\times 10^{4}\;\rm m/s} \end{align*} {/eq} 