At what speed would an object have to travel to be "in orbit" at ground level in a vacuum?

Answer 1

Set the 'centrifugal' force mv^2/r equal to the gravitational force GMm/r^2 and solve for v.

Answer 2

mathemati... is absolutely correct. But I thought I might provide a bit more on why the orbit can be determined from the weight of the object and the centripetal or equivalent centrifugal force.

First, centrifugal force acts outward along the radius of the orbit. It stems from one of Newton's Laws, which says...for every action (force) there is an equal and opposite reaction (force). Thus, while gravity is tugging inward, the centripetal force, there is an equal force pulling your object outward (in a straight line if it could according to another of Newt's Laws).

This outward reactive force is called centrifugal force and it can be calculated by the m v^2/r equation; where v is the tangential velocity of you object, r is the radius of the Earth, plus a smidge so you don't run into the trees and such.

So the inward pull from gravity (centripetal force) has to equal the outward pull from centrifugal force for the object to stay at the same r. In other words W = mg = m v^2/r near Earth's surface; so that v = sqrt(gr). Note the GmM/r^2 need not be used, it can be, but the weight W = mg is equivalent. Since g = 32.2 ft/sec^2 and r = 8,000 mi X 5,280 ft/mi, you can find the balancing velocity to be v = sqrt(32.2 X 8,000 X 5,280) in feet per second.

Answer 3

any orbit at this distance from the earth is not stable, unless of course, you really mean ground level, in which case, any speed that your vehicle can travel over land and water is the "orbital velocity"

Answer 4

You'd be travelling at roughly 17,500 mph and you would go round the earth once every 90 minutes.

Answer 5

You can do the math or simply go 7miles /sec which is escape velocity.

Answer 6

It would have to be damn level ground.

Answer 7

300000 km per second, the speed of light.