Orbital Motion under Gravity, Physics tutorial

Motion in a Vertical Circle:

1571_Motion in a Vertical Circle.jpg

The given represents the small body joined to the cord of length R and whirling in the vertical circle about a fixed point 0 to which other end of cord is attached. The motion although circular is not uniform as speed increases on way down and decreases on the way up. Forces on body at any point are its weight w=mg and tension T in cord. Resolving weight of the body in its components we have magnitude of normal component = w cos θ

Magnitude of tangential component = w sin θ

Resultant tangential and normal forces are:

F11 = w Sin Θ and F = T - w cos Θ

From Newton's second law then, we get tangential acceleration a11

a11 = F11/m g sin Θ

This is same as that of the body sliding down the frictional inclined

Plane of slope angle Θ

Normal radial acceleration a = V2/R is

a1 = F/m = T - w cos Θ/m = V2/R

T = m(V2/R + g cos Θ)

At the lowest point of path, Θ = O, thus Sin Θ = 0 and cos Θ = 1. So at this point F11 = 0 and a11 = 0 and acceleration is entirely radial (upward). Magnitude of tension, from Equation

is T = m(V2/R + g)

At highest point, Θ = 180° so Sin Θ = O and Cos Θ = - 1, and acceleration once more is entirely radial (downward). Tension for this case is

T = m(V2/r - g)

For this type of motion, there is certain critical speed VC at the highest point below which cord slacks and path will no longer be circular. To determine critical speed, we set T = 0 in Equation) i.e.

m(V2c/R - g) = 0

Therefore Vc= √Rg

Motion of a Satellite:

To launch the artificial satellite in space, multi-stage rockets are utilized. It raises satellite to the predetermined height, projects it in right direction with the velocity that allows it to revolve around Earth. This velocity is known as escape velocity. Escape velocity is a velocity with which a body must be projected to allow it to escape from gravitational influence of Earth.

We may calculate acceleration from velocity of satellite and the radius of the orbit therefore:

W = Fg = GMME/r2 = M(V2/r)............ Eq.1

From which we get

V2 = GME/r; V = √GME/r.................. Eq.2

We figure out from equation that larger the radius r, the smaller the orbital velocity.

We can also state speed of satellite in terms of the acceleration because of gravity g at surface of the earth that is provided by g = GME/R2. Combining this with Eq.2 we get

Since GME = gR2

V = √GME/r = √gR2/r

V = R√g/r.................. Eq.3

Acceleration given by aR = V2/r can also be expressed in terms of g thus:

aR = r2/R2g.................. Eq.4

Equation Eq.4 provides acceleration of gravity at radius r. Satellite, similar to any projectile is freely falling body. Acceleration is less than g at surface of earth in ratio of the square of radii.

Period T or time needed for one complete revolution is Equal to circumference of orbit divided by velocity, V:

T = 2Πr/v = (2Πr)/(R√g/r) = [(2Π)/(R/√g)] r3/2

It can be observe that longer the radius of the orbit the longer the period. R is a radius of earth here.

Parking Orbit:

The parking orbit is the temporary orbit utilized during launch of the satellite or other space probe. The launch vehicle boosts in parking orbit, and then coasts for the while, then fires again to enter final preferred trajectory. Alternative to the parking orbit is direct injection, where rocket fires continuously (apart from during staging) until its fuel is exhausted, ending with payload on final trajectory.

Reasons to use parking orbit:

  • It can amplify the launch window. For earth-escape missions, these are frequently fairly short (seconds to minutes) if no parking orbit is utilized. With the parking orbit, these can frequently be increased up to numerous hours.
  • For non-Low Earth orbit missions, preferred location for final burn may not be in the suitable spot. In particular, for earth-escape missions which want good northern coverage of trajectory, the right place for final burn is frequently in southern hemisphere.
  • For geostationary orbit missions, correct spot for final (or next to final) firing is usually on equator. In such a case, rocket is launched, coasts in the parking orbit until it is over equator, then fires again in the geostationary transfer orbit.
  • For manned lunar missions, the parking orbit permitted some checkout while still close to home, before committing to lunar trip.


The phenomenon of weightlessness takes place when there is no force of support on body. When body is efficiently in free fall, accelerating downward at acceleration of gravity, then you are not being supported. Sensation of evident weight comes from support that which feels from floor, from seat, etc. Different sensations of evident weight can take place on the roller-coaster or in aircraft as they can accelerate either upward or downward.

If you travel in the curved path in vertical plane, then when you go over top on such a path, there is essentially downward acceleration. Taking example of the roller-coaster that is constrained to follow a track, then the condition for weightlessness is met when the downward acceleration of the seat is equal to acceleration of gravity. Considering path of the roller-coaster to be the segment of a circle so that it can be associated to centripetal acceleration, situation for weightlessness is

Vweightless = from centripetal acceleration relationship v2/r = g

Weightlessness you may feel in the aircraft takes place any time aircraft is accelerating downward with acceleration 1g. It is possible to experience weightlessness for the significant length of time by turning nose of the craft upward and cutting power so that it travels in the ballistic trajectory. The ballistic trajectory is common kind of trajectory you get by throwing the rock or baseball, neglecting air friction. At every point on trajectory, acceleration is equal to g downward as there is no support.

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