CLASS 11 · CHAPTER 7 · MECHANICS

Gravitation

One inverse-square law explains a falling apple, a circular orbit, and a rocket leaving Earth forever. The only thing that changes between them is speed.

01 · See it

Watch it happen

Launch speed: 1.00× circular orbital speed

1.00

Launch speed (v/v꜀)

1.414×

Escape threshold, √2

Bound

Orbit type

Circular orbit

Drag the slider past 1.00× and the circle stretches into an ellipse. Keep going past 1.41× (that’s $\sqrt{2}$ ) and the satellite stops coming back at all, it has reached escape velocity.

02 · Derive it

Where the formula comes from

Newton’s law of gravitation between two masses:

F = \dfrac{GMm}{r^2}

For a circular orbit, this force alone has to supply the centripetal force, nothing else is pulling the satellite inward:

\dfrac{GMm}{r^2} = \dfrac{mv^2}{r} \;\;\Rightarrow\;\; \textcolor{#e08a1e}{v_{\text{orbital}}} = \sqrt{\dfrac{GM}{r}}

Escape velocity comes from energy instead of force: launch with just enough kinetic energy to cancel the negative gravitational potential energy at infinity, $\tfrac{1}{2}mv_e^2 = \dfrac{GMm}{r}$ :

Escape velocity

\textcolor{#e08a1e}{v_e} = \sqrt{\dfrac{2GM}{r}} = \sqrt{2}\,v_{\text{orbital}}

03 · Break it

Where the shortcut stops working

“g decreases as you move away from Earth’s surface” is true whether you go up or down, which is exactly why it’s easy to assume both directions obey the same formula. They don’t. Above the surface, all of Earth’s mass still pulls on you, so $g$ follows the full inverse-square law:

g(h) = \dfrac{GM}{(R+h)^2} \approx g_0\left(1 - \dfrac{2h}{R}\right) \quad \text{for } h \ll R

Below the surface, only the mass inside your radius still counts, the shell of Earth above you contributes nothing (by the shell theorem), so the mass pulling on you itself shrinks as you descend. The result is a straight line, not a curve:

g(d) = g_0\left(1 - \dfrac{d}{R}\right)

Same surface, same “g decreases,” but a different law in each direction, linear inside, inverse-square outside. Mixing up the two formulas is one of the most common errors in this chapter.

04 · Master it

Apply it under exam conditions

Q1. Find the escape velocity from Earth’s surface. (g = 9.8 m/s², R = 6.4 × 10⁶ m)

Using $v_e = \sqrt{2gR}$ (since $GM = gR^2$ at the surface):

v_e = \sqrt{2\times9.8\times6.4\times10^6} \approx \textcolor{#e08a1e}{11.2\text{ km/s}}

Q2. At what depth below the surface does g fall to the same value it has at a height equal to R/10 above the surface?

At height $h = R/10$ : $g(h) \approx g_0(1 - 0.2) = 0.8g_0$ . Setting the depth formula equal: $g_0(1 - d/R) = 0.8g_0 \Rightarrow d = \textcolor{#e08a1e}{0.2R}$ , twice as deep as the height was high, because the two laws have different slopes near the surface.

05 · FAQs

Quick answers

Why don't satellites fall down?+

They are falling, continuously, but they're also moving sideways fast enough that the curve of their fall matches the curve of the Earth beneath them, so they keep missing the ground.

What's the difference between orbital velocity and escape velocity?+

Orbital velocity is exactly enough speed to keep circling at a given radius; escape velocity is √2 times larger, enough to leave and never come back at all.

Does g change the same way above and below the surface?+

No. Above the surface it falls off as an inverse square; below it, only the enclosed mass counts, so it falls off linearly. They're genuinely different laws that happen to agree at the surface itself.

Why doesn't the Moon go flying off into space?+

Because Earth's gravity supplies exactly the centripetal force needed to keep it curving around us. There's no separate outward force fighting gravity, just gravity providing the turn.

Is weightlessness in orbit the absence of gravity?+

No, gravity at typical orbital altitudes is nearly as strong as at the surface. Weightlessness happens because the spacecraft and everything inside it are falling together at the same rate, so nothing pushes against anything else.

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