What is the maximum circular speed the astronaut can have before he breaks free of gravity and goes drifting off into space?
One does not drift in space, gravity extends forever, and one is always under its influence. There is a certain velocity call escape velocity, which means that one can leave orbit and the force of gravity will not make it return, but the object is not "free" of gravity, it still slows him down.
I don't know what you mean.
The escape velocity for any body may be calculated from Ve = sqrt[2µ/r] where Ve = the escape velocity in feet per second, µ = the gravitational constant of the body (1.407974x10^16 ft.^3/sec.^2 for Earth) and r = the surface distance in feet. For earth, with an equatorial radius of ~3963 miles r becomes 3963(5280) = ~20,924,640 feet and Ve = sqrt[2(1.407974x10^16)/20,924,640] = 36,685 fps or 25,000 mph.
As Bob Pursley pointed out, an astronaut/spacecraft given this minimal escape velocity will end up on a parabolic trajectory which will ultimately take it to the edge of the Earth's sphere of influence (typically assumed to be ~575,000 miles) with zero velocity, where it will remain forever. The astronaut/spacecraft wil feel the pull of earth's gravity for the entire trip, ultimately reducing its velocity to essentially zero.
Any velocity less than escape velocity will place the astronaut/spacecraft on a highly eccentric elliptical orbit with a very large apogee, only to return to the Earth's vicinity at a later date.
To determine the maximum circular speed an astronaut can have before breaking free of gravity and drifting off into space, we need to consider the concept of escape velocity.
Escape velocity is the minimum velocity needed for an object to escape the gravitational pull of a planet or any celestial body. It is calculated using the formula:
v = √(2 * G * M / r)
Where:
- v is the escape velocity
- G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2)
- M is the mass of the celestial body (in this case, Earth)
- r is the distance between the center of the celestial body and the object (in this case, the radius of Earth)
For Earth, the average radius is approximately 6,371 kilometers (6,371,000 meters) and the mass is approximately 5.972 × 10^24 kilograms.
Now, plugging those values into the formula:
v = √(2 * 6.67430 × 10^-11 * 5.972 × 10^24 / 6,371,000)
After calculating this expression, we find that the escape velocity from Earth is approximately 11,186 meters per second (or 40,270 kilometers per hour). So, any object or astronaut traveling at a speed greater than this will overcome Earth's gravitational pull and drift off into space.