Irodov 1.222: How Orbital Speed Sets the Altitude – Amit & Deepali Sinha Foundation

Problem Irodov 1.222

A satellite moves in a circular orbit about the Earth with speed $v$ , measured in a frame whose axes keep a fixed orientation in space and translate along with the Earth’s center, without sharing the Earth’s own rotation. Given the Earth’s radius $R$ and the free-fall acceleration $g$ at its surface, find the satellite’s height $h$ above the surface.

The satellite rides a circular orbit of radius r=R+h; its velocity v is tangent to the orbit, and h is the altitude dimensioned out from the Earth's surface to the ring itself. — The satellite rides a circular orbit of radius $r=R+h$ ; its velocity $\vect v$ is tangent to the orbit, and $h$ is the altitude dimensioned out from the Earth’s surface to the ring itself.

Given

Find

h

Solution

Step 1Circular orbit condition fixes the radius

Gravity supplies exactly the centripetal force for a circular orbit of radius $r$ about the Earth’s center, with $M$ the Earth’s mass; equating the two forms of that force and canceling the satellite’s mass and one power of $r$ gives $r$ directly in terms of $v$ and $\gamma M$ .

\begin{aligned} \frac{\gamma M m}{r^2} &= \frac{mv^2}{r} \\ \gamma M &= v^2 r \\ r &= \frac{\gamma M}{v^2} \end{aligned} \tag{1}

Step 2Write

\gamma M

through the surface value of

g

g = \frac{\gamma M}{R^2} \;\Longrightarrow\; \gamma M = gR^2 \tag{2}

Step 3Height above the surface

The satellite’s height above the surface is the orbital radius less the Earth’s own radius; substituting (2) into the orbital radius (1) and factoring out $R$ gives $h$ directly.

\begin{align*}r &= \frac{gR^2}{v^2} \\ h &= r-R \\ &= \frac{gR^2}{v^2}-R\end{align*}

h = R\left(\frac{gR}{v^2}-1\right)

(3)

Check

As $v\to\sqrt{gR}$ , the speed of a circular orbit grazing the surface itself, $h\to0$ : the formula correctly places the satellite back on the ground. As $v$ decreases below $\sqrt{gR}$ , $h$ grows without bound, consistent with the general fact that a circular orbit’s speed falls as the orbit gets larger ( $v\propto1/\sqrt r$ ), so a slower launch speed can only be matched by a farther-out orbit.

Remark

The problem’s careful phrasing, “velocity relative to the frame moving translationally... fixed to the Earth’s rotation axis,” rules out the Earth’s own spin: that frame’s axes do not turn, so $v$ enters Newton’s second law directly as the ordinary orbital speed with no correction for the ground moving underneath the satellite. A satellite’s speed measured instead relative to the rotating, Earth-fixed frame would need $v$ combined with $\vect\Omega\times\vect r$ , the velocity of the point of the co-rotating frame that instantaneously coincides with the satellite; for an orbit confined to the equatorial plane this reduces to the scalar $\Omega r$ , evaluated at the satellite’s own orbital radius $r$ , which is larger than the local speed of the ground itself, $\Omega R$ . This is exactly the correction that a retrograde or a prograde satellite tracked from the Earth’s surface picks up and this one is built to avoid.