\[
\mathbf g = \frac{\mathbf F{\text{test}}}{m{\text{test}}}
\[
\mathbf F = -\frac{GMm}{r^{2}}\;\hat{\mathbf r}
\]
where r is the separation of the centres and \hat r points from the source to the test mass.
\[
\boxed{\mathbf g(r)= -\frac{GM}{r^{2}}\;\hat{\mathbf r}}
\]
The field is radial, points toward the source (negative sign), and its magnitude falls as 1/r².
When both masses are free to move, the mutual attractive force is
\[
\boxed{\mathbf F_{12}= -\frac{GMm}{r^{2}}\;\hat{\mathbf r}}
\]
Both bodies experience forces of equal magnitude and opposite direction (Newton’s 3rd law). This expression is a required item of the syllabus (Section 13.2).
\[
\phi(r)=\frac{U{\text{grav}}(r)}{m{\text{test}}}\qquad\bigl[\text{J kg}^{-1}\bigr]
\]
\[
\phi(r)= -\int_{\infty}^{r}\mathbf g\cdot d\mathbf r'
= -\int_{\infty}^{r}\frac{GM}{r^{\prime 2}}\,dr^{\prime}
= -\frac{GM}{r}
\]
\[
\boxed{\phi(r)= -\frac{GM}{r}}\qquad\text{(J kg}^{-1}\text{)}
\]
\[
\mathbf g = -\nabla\phi \;\; \text{or (radial)}\;\; g(r)= -\frac{d\phi}{dr}
\]
Substituting φ = -GM/r reproduces the field g = -GM/r², confirming consistency.
For a solid sphere of mass M and radius R:
\[
\phi_{\text{out}}(r)= -\frac{GM}{r},\qquad
g_{\text{out}}(r)= -\frac{GM}{r^{2}}
\]
\[
g_{\text{in}}(r)= -\frac{GM}{R^{3}}\,r
\]
Integrating from R to r gives the potential:
\[
\boxed{\phi_{\text{in}}(r)= -\frac{GM}{2R^{3}}\bigl(3R^{2}-r^{2}\bigr)}
\]
(Both results are part of the syllabus, Section 13.4.)
\[
\boxed{U_{\text{grav}} = -\frac{GMm}{r}}\qquad\text{(J)}
\]
This follows directly from U = m\phi or from the work‑integral method.
| Situation | Key equation | Result (after substitution) |
|---|---|---|
| Circular orbit of radius r | \[ E_{\text{tot}} = K + U = \tfrac12 mv^{2} - \frac{GMm}{r} \] with centripetal condition v^{2}=GM/r | \[ E_{\text{tot}} = -\frac{GMm}{2r} \] |
| Escape speed from a planet of radius R | \[ \tfrac12 mv_{\text{esc}}^{2}+U(R)=0 \] | \[ v_{\text{esc}} = \sqrt{\frac{2GM}{R}} \] |
| Energy change when a satellite moves from r{1} to r{2} | \[ \Delta U = U(r{2})-U(r{1}) = -GMm\!\left(\frac{1}{r{2}}-\frac{1}{r{1}}\right) \] | Positive ΔU ⇒ work must be done against gravity. |
Find the total energy of a 500 kg satellite in a circular orbit 300 km above the Earth’s surface.
\[
v = \sqrt{\frac{GM_{\oplus}}{r}}
= \sqrt{\frac{(6.674\times10^{-11})(5.97\times10^{24})}{6.67\times10^{6}}}
\approx 7.73\times10^{3}\,\text{m s}^{-1}
\]
\[
E = K+U = -\frac{GM_{\oplus}m}{2r}\approx -1.5\times10^{10}\,\text{J}
\]
| Symbol | Quantity | SI unit | Typical value |
|---|---|---|---|
| G | Universal gravitational constant | m³ kg⁻¹ s⁻² | 6.674 × 10⁻¹¹ |
| M, m | Masses of interacting bodies | kg | varies |
| R | Radius of a spherical body (e.g., Earth) | m | varies |
| r | Separation between the centres of two bodies | m | varies |
| g | Gravitational field (acceleration due to gravity) | m s⁻² | ≈9.81 at Earth’s surface |
| φ | Gravitational potential (per unit mass) | J kg⁻¹ | negative |
| U | Gravitational potential energy of a system | J | negative |
• Gravitational field: \mathbf g = -GM/r^{2}\,\hat r (radial, inward).
• Gravitational force between two point masses: \mathbf F = -GMm/r^{2}\,\hat r.
• Gravitational potential: \phi = -GM/r (J kg⁻¹).
• Gravitational potential energy of a two‑mass system: U = -GMm/r (J).
• Outside a uniform sphere the same formulas apply; inside the field varies linearly and the potential is \phi_{\text{in}} = -\dfrac{GM}{2R^{3}}(3R^{2}-r^{2}).
• These relationships underpin A‑Level questions on satellite motion, escape speed, and energy changes in gravitational fields.
Your generous donation helps us continue providing free Cambridge IGCSE & A-Level resources, past papers, syllabus notes, revision questions, and high-quality online tutoring to students across Kenya.