# How fast would a rocket have to go to maintain in the earth orbit without falling down? And how frequent would rocket boosters be needed? [closed]

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?

In order to maintain a circular orbit, a rocket needs to experience a force (called the centripetal force) equal to $$F = \frac{mv^2}{r},$$ 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 $$F = \frac{mv^2}{r} = G\frac{Mm}{r^2}$$ $$v^2 = \frac{GM}{r}$$ $$v = \sqrt\frac{GM}{r}$$ 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 $$5.9722 \times 10^{24}\ \textrm{kg}$$. 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 $$v = \sqrt\frac{(6.674 \times 10^{−11})(5.9722 \times 10^{24})}{6771000} = 7\,670\ \textrm{m/s} = 27\,600\ \textrm{km/hr}$$

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.

• The ISS Orbits at a height of about 400km, a fair bit higher than the startlingly low 400m you have quoted. I dare say at that altitude it wouldn't stay in orbit (or in one piece) for very long. I would love to watch that go by at mach 23, from a safe distance of course. Apr 20 at 5:13
• @Turksarama It is true that there is a slight difference between meters and kilometers. Fixed. Apr 20 at 6:04

You can calculate this by assuming the spacecraft is in a circular orbit and the centripetal force equals the gravitational force of the Earth. So $$F = GMm/r^2 = mv^2/r$$. $$m$$ cancels out, $$G$$ is a constant, $$M$$ is the mass of the Earth and $$r$$ is the radius of your orbit. Solving the equation gives you $$v$$.

How frequently boosters are needed is a more difficult question since it depends on the air resistance at your given orbit. The easy way out is just to see what the ISS currently does, and a quick search finds this related question with the answer.

We have equations to describe the necessary velocity for specific masses to orbit another mass based on how far away the satellite is. The equation for orbital velocity can be written in terms of an orbital period, or gravitational constant and mass, assuming an elliptic orbit with small eccentricity (nearly circular orbit).

$$v \approx \sqrt{\frac{2\pi a}{T}} \approx \sqrt{\frac{\mu}{a}}$$

In the first approximation, orbital velocity depends on the satellite orbital period $$T$$, and the semi-major axis of its elliptic orbit $$a$$ (Think distance $$r$$ for circular orbit). In the second approximation, it depends on the standard gravitational parameter ($$\mu = GM$$), and again the semi major-axis $$a$$. For this form of $$\mu$$, it is assumed that the satellite mass $$m$$ is much less than the orbited mass $$M$$.

So for your question, how fast a rocket must travel to maintain orbit - the answer is that if its mass is much less than the Earth, then it depends only on how far away from the Earth it is. A good approximation is thus $$v \approx \sqrt{\frac{GM}{r}}$$

So for an object orbiting at 400km above the Earth, distance $$r$$ is $$400km+6371km=6771km$$ (distance above surface plus Earth radius) and the orbital velocity should be approximately $$v \approx \sqrt{\frac{(6.67\times10^{-11})(5.97\times10^{24})}{6771000}} \approx \space 7669 \space m/s \approx \space 27607 \space km/h$$

Orbits will naturally decay and spiral towards the Earth, unless as you mentioned an orbital burn is periodically performed to maintain the orbital speed and path. This is commonly due to atmospheric drag for low Earth orbits where there are some trace of molecules acting to slow the satellite. To find how often a correction must be done, you must calculate how the orbit is decaying over time by looking at the area of the satellite body and the density of the atmosphere causing drag.