Sincere Garcia

2022-10-08

I was thinking how fast would a rocket have to go to maintain in the in earth orbit? I was thinking of the ISS going 7000kph with rocket boosters every 6 months. What is the most efficient way to solve this?

graulhavav9

Beginner2022-10-09Added 14 answers

In order to maintain a circular orbit, a rocket needs to experience a force (called the centripetal force) equal to

$F=\frac{m{v}^{2}}{r},$

where m is the mass of the rocket, v is its velocity, and r is the radius of its orbit. This force is provided by gravity, which is equal to

$F=\frac{GMm}{{r}^{2}},$

where G is the gravitational constant ($6.674\times {10}^{-11}$ when using metric units), M is the mass of the gravitating body (Earth, in this case), and r is the distance from the center of the body. Setting these equal to each other, we get

$F=\frac{m{v}^{2}}{r}=G\frac{Mm}{{r}^{2}}$

${v}^{2}=\frac{GM}{r}$

$v=\sqrt{\frac{GM}{r}}$

Notice that the mass of the rocket does not factor into this final equation. So, now we can substitute some numbers. The mass of Earth, M, is $5.9722\times {10}^{24}\text{}\text{kg}$. The ISS orbits at a height of 400 kilometers above the Earth's surface, which is 6371 km in radius, which gives r as 6771000 meters. Plugging all this in, we get

$v=\sqrt{\frac{(6.674\times {10}^{-11})(5.9722\times {10}^{24})}{6771000}}=7\phantom{\rule{thinmathspace}{0ex}}670\text{}\text{m/s}=27\phantom{\rule{thinmathspace}{0ex}}600\text{}\text{km/hr}$

Once an object has this speed at that height, it will stay in orbit with no need for rockets or other propulsion. The reason that the space station has to fire its rockets is because of Earth's atmosphere. The very thin air at the space station's height creates a little bit of drag that causes it to slow down and lose altitude.

$F=\frac{m{v}^{2}}{r},$

where m is the mass of the rocket, v is its velocity, and r is the radius of its orbit. This force is provided by gravity, which is equal to

$F=\frac{GMm}{{r}^{2}},$

where G is the gravitational constant ($6.674\times {10}^{-11}$ when using metric units), M is the mass of the gravitating body (Earth, in this case), and r is the distance from the center of the body. Setting these equal to each other, we get

$F=\frac{m{v}^{2}}{r}=G\frac{Mm}{{r}^{2}}$

${v}^{2}=\frac{GM}{r}$

$v=\sqrt{\frac{GM}{r}}$

Notice that the mass of the rocket does not factor into this final equation. So, now we can substitute some numbers. The mass of Earth, M, is $5.9722\times {10}^{24}\text{}\text{kg}$. The ISS orbits at a height of 400 kilometers above the Earth's surface, which is 6371 km in radius, which gives r as 6771000 meters. Plugging all this in, we get

$v=\sqrt{\frac{(6.674\times {10}^{-11})(5.9722\times {10}^{24})}{6771000}}=7\phantom{\rule{thinmathspace}{0ex}}670\text{}\text{m/s}=27\phantom{\rule{thinmathspace}{0ex}}600\text{}\text{km/hr}$

Once an object has this speed at that height, it will stay in orbit with no need for rockets or other propulsion. The reason that the space station has to fire its rockets is because of Earth's atmosphere. The very thin air at the space station's height creates a little bit of drag that causes it to slow down and lose altitude.

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