Gravitational Force Between Point Masses
In physics, the gravitational force between two point masses is described by Newton’s law of universal gravitation.
Think of it like a cosmic handshake: the more massive the objects, the stronger the handshake, and the farther apart they are, the weaker it becomes. 🤝
1️⃣ The Gravitational Force Formula
\$F = G\frac{m1m2}{r^2}\$
where:
- \$F\$ = gravitational force (N)
- \$G\$ = gravitational constant \$6.674\times10^{-11}\,\text{Nm}^2\text{/kg}^2\$
- \$m1, m2\$ = masses (kg)
- \$r\$ = distance between centres (m)
2️⃣ Centripetal Acceleration
For an object moving in a circle of radius \$r\$ at speed \$v\$, the required centripetal acceleration is
\$a_c = \frac{v^2}{r}\$
The gravitational force provides this acceleration when the object orbits a massive body. 🌍
3️⃣ Circular Orbit Condition
In a stable circular orbit, the gravitational force equals the centripetal force:
\$G\frac{M m}{r^2} = m\frac{v^2}{r}\$
Simplifying (cancel \$m\$):
\$v^2 = \frac{GM}{r}\$
So the orbital speed depends only on the mass of the central body (\$M\$) and the orbital radius (\$r\$).
🔄 If you double the radius, the speed drops to about \$1/\sqrt{2}\$ of its previous value.
4️⃣ Example: Earth‑Moon System
Let’s calculate the Moon’s orbital speed around Earth.
- \$M_{\text{Earth}} = 5.97\times10^{24}\,\text{kg}\$
- \$r_{\text{Earth–Moon}} = 3.84\times10^8\,\text{m}\$
Using \$v = \sqrt{GM/r}\$:
\$v = \sqrt{\frac{6.674\times10^{-11}\times5.97\times10^{24}}{3.84\times10^8}} \approx 1.02\times10^3\,\text{m/s}\$
So the Moon travels about 1 km/s around Earth. 🚀
5️⃣ Exam Tips Box
📝 Key Points to Remember:
- Always write the full equation \$F = Gm1m2/r^2\$ before simplifying.
- When cancelling the orbiting mass \$m\$, note that it disappears from the final speed expression.
- Check units: \$G\$ is in \$\text{Nm}^2\text{/kg}^2\$, so \$v\$ will be in \$\text{m/s}\$.
- For circular orbits, remember \$v^2 = GM/r\$; for elliptical orbits, use the vis‑viva equation.
- Use the “handshake” analogy to explain why more massive bodies pull harder.
6️⃣ Quick Practice Problems
- Calculate the orbital speed of a satellite at an altitude of 300 km above Earth’s surface.
- If a planet has twice Earth’s mass but the same radius, how does the orbital speed at its surface change?
- Show that the gravitational force between two equal masses at a distance of 10 m is 1/100 of the force when they are 1 m apart.
7️⃣ Summary Box
Remember:
- Gravitational force decreases with the square of distance.
- In a circular orbit, the required centripetal acceleration is supplied by gravity.
- Orbital speed depends only on the central mass and orbital radius.