gravity in orbit?

Question

astranauts who go into orbit are still under the influence of gravity. i was wondering what the typical gravitational field strength experinced by astranauts is?

Answer 1

I do this example in my physics class every year - I have the students calculate the value for g (9.8 m/s2 on Earth's surface) at the height the Space Shuttle typically orbits at, about 300 km above Earth's surface. Using that value, g = 8.97 m/s2, or 91.5% of the surface value. So the decrease is not much, roughly 10% less on orbit than on the surface.

You can do the calculation yourself, if you want. Here are the equation and the numbers:
g = GM/r^2
G = 6.67x10^-11
M = mass of Earth = 5.98x10^24 kg
r = radius of Earth + 300km = 6.67x10^6 m

Answer 2

It's called micro-gravity BUT it's not that the gravitational field from the earth is small !
[warning! science content] When something is moving around in a circle it needs to be held back by something. There is a [fictitious] force, called the centrifugal force that pushes the object away from the center of rotation.
An object is in orbit (by definition) when this centrifugal force is exactly cancelled by the gravitational pull. Therefore, it's not that there's no gravity when in orbit, it's that the gravitational field is cancelled.
[as a matter of fact, if it wasn't for the gravitational pull of the sun, we would fly off the earth at night]
So, why "microgravity". Because for a given rotational speed (simplifying here to circular orbits) the gravitational force and the centrifugal force cancel each other exactly _only_at_one_point_ If you move away from that point, you start feeling a force. So, for example, when the space shuttle is in orbit, only one point of the shuttle (the center of mass) would feel zero force.
[just to keep going... this "force" is really a tidal force... it goes as 1 over r^4 so the lower the orbit, the higher the force. It's also the force that would kill you when you tried to fly through a wormhole and ... damn, I could keep going but I'm sure everyone stopped reading by now...]

-jose-

Answer 3

You have to be a bit more precise in your question. Most space missions are only 1-200 miles above the earth, so the strength of the gravitation force from the earth is only about 1% less than what you or I feel. The difference is that the astronaut (and the spaceship) are in free-fall so the astronaut doesn't *feel* that force of gravity.

On the other hand, since there is a very slight difference in how far away one side of the sapacecraft is from the earth than the other side, there are very small tidal forces that do have to be taking into account. This is why we say an astronaut is in micro-gravity. The size of those tidal forces is around one *millionth* of the gravitational force we feel on earth.

Answer 4

In orbit, an astronaut is in free fall and so will experience Zero G
If the astronaut was stationary, ie not orbiting, he would be subject to the inverse square law of gratitational attraction
Gravity on earth is about 9.8 m/s².
I have seen suggestions for gravitational effect on satellites at anywhere between 2 m/s² and 8 m/s² (remember there is a differnece between the height of a geostationary orbiting satellite, and a non-geostationary satellite)

Answer 5

The gravity is about the same for low orbit as it is on the ground. If they were going that fast in a tunnel a mile under the earth they'd still be floating.

Answer 6

Interesting.The question boils down to finding out how much of Earth's gravity is still there in the environs of a space ship.We will attempt to calculate it. We make the following assumptions.
The launch is equatorial and space ship stabilises in a geosynchronous orbit.we have the basic equation
GMm/r^2=mg where m is the mass of the space ship,M mass of Earth, r is the distance from centre of Earth to space ship,G the universal constant and g acceration due to gravity.
g=GM/r^2
G=6.6742*10^-11Nm^2kg^-2
M=2*10^24kg
r=Equatorial radius+distance to space ship
=3963.35mi+22236mi(wikipedia)
=26199.35
=26200mi=26200*1600meters
g=6.672*10^-11*6*10^24
/[26200*1600]^2
=6.6742*6*10^-11*10^24
/1.757*10^15
=.2279M/sec^2
Reduction in acceleration due to gravity
as compared to ground level
=[9.81-.2279]/9.81
=9.5721/9.81=.975
There is reduction of 97.5%

Answer 7

gravity field is less but the gravitational force is used for centripetal force

Answer 8

It is from the earth or either the solar system. each planets in our solar system have their own grativy forces which will influence anything in the space.

Answer 9

depends on if its a low orbit or high. gravity is stronger the closser you are.

Answer 10

Not much