Nilesat 201 is a communications
satellite that orbits Earth at a radius of 35,800 kilometers. What is the orbital speed of
Nilesat 201? Assume that the satellite follows a
circular orbit. Use a value of 5.97 times 10 to the
24 kilograms for the mass of Earth and 6.67 times 10 to the negative 11 meters cubed
per kilogram second squared for the universal gravitational constant. Give your answer in scientific
notation to two decimal places.
Here, we want to find the orbital
speed of a satellite in the special case of circular orbit. So we’ll use the orbital speed
formula 𝑉 equals the square root of 𝐺 times 𝑀 divided by 𝑟, where 𝑉 is orbital
speed. 𝐺 is the universal gravitational
constant. 𝑀 is the mass of the large body
being orbited. Here, that’s Earth. And 𝑟 is orbital radius, which
extends between the Earth’s and the satellite’s centers of gravity.
We’ve been given values for 𝐺, 𝑀,
and 𝑟. But before we substitute them into
the formula, all of these terms should be expressed in base SI units. 𝐺 is expressed in meters,
kilograms, and seconds, so it’s good to go, as well as mass in kilograms. But the kilometer is not the base
SI unit of distance, so we’ll need to convert it into plain meters. Recall that one kilometer is equal
to 1,000 meters. So we’ll multiply 𝑟 by this
conversion factor: 1,000 meters divided by one kilometer, which is just equal to
one. And we can cancel units of
kilometers. So now we have 𝑟 equals 35,800,000
meters. The units are correct, but let’s
write this in scientific notation.
𝑟 equals 3.58 times 10 to the
seven meters. Now, copying the formula below,
let’s substitute in these values. 𝑉 equals the square root of 6.67
times 10 to the negative 11 meters cubed per kilogram second squared times 5.97
times 10 to the 24 kilograms divided by 3.58 times 10 to the seven meters, which
comes out to about 3,335 meters per second. Finally, in scientific notation, to
two decimal places, we have found that the orbital speed of Nilesat 201 is 3.34
times 10 to the three meters per second.