# Free fall around Earth

Is it a necessary condition that a satellite must be given an initial perpendicular velocity so as to make it orbit around the Earth? Will it go into orbit if dropped with zero initial velocity?

• So the satellite is just popped into space, with no velocity, somewhat near Earth? – Dave Mar 10 '17 at 3:19
• Perpendicular to what? Perpendicular is not a uniquely-defined direction. – Bill N Mar 10 '17 at 3:31
• There's a range of initial velocities that orbit, and a range that have orbits that intersect the Earth's surface, and a range of velocities that just... vanish off into the distance. – Whit3rd Mar 10 '17 at 4:33
• This answer tells you why you need a tangential component of velocity for something to orbit: physics.stackexchange.com/questions/312492/… – Yashas Mar 11 '17 at 6:05

For an earth/satellite system (i.e. ignoring any other bodies), there is only one force: the gravitational force between earth and satellite. This force is always directed between the centres of mass of the bodies. In this case that means it is directed towards the earth's centre.

If a satellite has no tangential motion (if you just "place" it at a point near the earth) then gravity will cause it to accelerate directly towards the centre of the earth. In other words, it will fall straight down to earth.

To make the satellite stay in orbit, you must give it sufficient horizontal velocity so it always "falls past the edge of the earth". The example that Newton used is a cannon. When we fire it horizontally, the cannonball moves at a fixed horizontal speed (ignoring air friction). At the same time it accelerates towards the earth's centre. The result is that it travels along a parabola until it hits the ground.

If we use more gunpowder, the ball will go faster and fly further. As we keep shooting it faster and faster, eventually we will get to a speed of about $7.7 km/s$ at which point the ball will fly so far that it goes right around the earth. The ball is then "in orbit" and can keep flying around forever - always ignoring air resistance.

To recapitulate, without horizontal velocity the satellite will fall straight down. The faster it moves horizontally, the further it will fly before it hits. Only at $7.7 km/s$ (orbital speed) or higher will it stay in orbit.

If the speed increases above $11.2 km/s$ then earth's gravity is no longer able to keep the satellite in orbit, and it will fly away. It has reached "escape speed".

• although if you did pop it in near enough to the moon, the horizontal acceleration could come from the moon's gravity. Just to confuse the issue. its not a great orbit though, very high, and may be perturbed by the moon later. – JMLCarter Mar 10 '17 at 9:15
• @JMLCarter That's why I said "ignoring any other bodies". – hdhondt Mar 10 '17 at 9:34

You can think of a orbiting satellite as a body performing uniform circular motion around the Earth. For this to happen, the gravitational force due to the Earth itself must be the centripetal force that is needed to sustain the circular motion: \begin{equation} F=m\frac{v^2}{r}=G\frac{mM}{r^2}, \end{equation} where $m$ is the mass of the satellite and $v$ its speed, $r$ the distance from the center of the Earth, $G$ is Newton's constant and $M$ the mass of the Earth. So, as you can see, you need a specific speed to have a satellite orbiting the Earth at a given distance from its center.

On the other hand, if the body has no initial velocity, it will simply fall in the direction of the force, i.e. towards the center of the Earth.