# 15.2 Gravitational Field Strength How does the Sun keep Earth in orbit from millions of miles away? The answer is that massive objects create an invisible area of influence around them called a **gravitational field**. This section explains how we measure the pull of this field at any given point, a value we call **gravitational field strength**, or more simply, 'g'. We explore why 'g' is about 9.8 m/s² on Earth's surface and how its strength gets weaker the higher you go. - **Gravitational Field:** The region around a massive object where its gravitational force is exerted. - Represented by **field lines** that indicate the direction and strength of gravity. - **Closer lines** signify a stronger field; **farther apart lines** indicate a weaker field. - For a point mass, field lines point radially inward uniformly in all directions, forming a symmetric pattern. - **Gravitational Field Strength (g):** - A **vector quantity** with a magnitude 'g' pointing in the direction of the gravitational force. - Defined as the gravitational force per unit mass at a given point: $g = F_g / m$. - **Equivalence:** Gravitational field strength 'g' is equivalent to the acceleration due to gravity, $a_g = F_g / m = g$. ## Value of 'g' on the Surface of the Earth Derived from Newton's law of universal gravitation and Newton's second law: $$F_g = G \frac{m_0 M_E}{R_E^2} \quad \text{and} \quad F_g = m_0 g$$ Equating these gives: $$g = G \frac{M_E}{R_E^2}$$ Using $M_E = 6 \times 10^{24} \, \mathrm{kg}$, $R_E = 6.4 \times 10^{6} \, \mathrm{m}$, and $G = 6.67 \times 10^{-11} \, \mathrm{N \, m^2 \, kg^{-2}}$, the value of 'g' on Earth's surface is approximately $9.8 \, \mathrm{m \, s^{-2}}$. **Key dependencies:** The value of 'g' depends only on the mass ($M_E$) and radius ($R_E$) of the planet, not on the mass of the object ($m_0$). **General form for any planet:** $g = G \frac{m}{r^2}$, where 'm' is the planet's mass and 'r' is its radius. ## Variation of 'g' with Altitude The value of 'g' changes as distance from the Earth's center changes. At a height 'h' above the Earth's surface, the gravitational field strength $g_h$ is given by: $$g_h = G \frac{M_E}{(R_E + h)^2}$$ This can also be expressed using the surface gravity 'g': $$g_h = \frac{g R_E^2}{(R_E + h)^2}$$ **Effect:** As altitude 'h' increases (moving away from the Earth's surface), the value of 'g' decreases. **Table 15.1: Variation in Gravitational Acceleration with Altitude** | Altitude (km) | $g_h$ (m s⁻²) | Example | |---------------|---------------|---------| | 0 | 9.8 | Average Earth radius | | 8.8 | 9.8 | Mount Everest | | 36.6 | 9.7 | Highest manned balloon | | 400 | 8.7 | Space shuttle orbit | | 35,700 | 0.2 | Communication satellites | **Why 'g' is approximately constant for small changes in height near Earth's surface:** For small heights $h \ll R_E$, the change in 'g' is negligible because Earth's radius ($R_E = 6.4 \times 10^6$ m) is very large compared to typical terrestrial heights. The fractional change is approximately $2h/R_E$, which for heights of a few kilometers is less than 0.1%. This explains why the acceleration due to gravity is taken as constant (9.8 m/s²) for everyday practical purposes near Earth's surface. ## Gravitational Field Lines for a Uniform Sphere Earth is considered a perfectly uniform sphere for easier analysis. **Outside the Sphere:** The field behaves as if all the sphere's mass is concentrated at its center, similar to a point mass. The field strength reduces rapidly with distance (inverse-square law). **Inside the Sphere:** The gravitational field strength increases steadily toward the surface, as only the mass within the radius to the point affects the field. ## Possible Questions and Answers **Q:** What is the gravitational field strength at 120 km above the Earth's surface? **A:** Using $g_h = \frac{g R_E^2}{(R_E + h)^2}$ with $g = 9.8 \, \mathrm{m \, s^{-2}}$, $R_E = 6.4 \times 10^6 \, \mathrm{m}$, and $h = 0.12 \times 10^6 \, \mathrm{m}$: $$g_h = \frac{9.8 \times (6.4 \times 10^6)^2}{(6.4 \times 10^6 + 0.12 \times 10^6)^2} = 9.4 \, \mathrm{m \, s^{-2}}$$ **Q:** Determine the free-fall acceleration at the surface of Jupiter, given its mass $m_J = 1.898 \times 10^{27} \, \mathrm{kg}$ and radius $r_J = 7.15 \times 10^7 \, \mathrm{m}$. **A:** Using $g_J = G \frac{m_J}{r_J^2}$: $$g_J = 6.67 \times 10^{-11} \times \frac{1.898 \times 10^{27}}{(7.15 \times 10^7)^2} = 24.77 \, \mathrm{m \, s^{-2}}$$ **Q:** Calculate the weight of a 60 kg astronaut on Jupiter's surface. **A:** Using $W = m g_J$: $$W = 60 \, \mathrm{kg} \times 24.77 \, \mathrm{m \, s^{-2}} = 1486.2 \, \mathrm{N}$$ **Q:** What will be the value of gravitational field strength 'g' at $35,700 \, \mathrm{km}$ (geostationary orbit) above Earth's surface? **A:** Using $g_h = \frac{g R_E^2}{(R_E + h)^2}$ with $g = 9.8 \, \mathrm{m \, s^{-2}}$, $R_E = 6.4 \times 10^6 \, \mathrm{m}$, and $h = 35.7 \times 10^6 \, \mathrm{m}$: $$g_h = \frac{9.8 \times (6.4 \times 10^6)^2}{(6.4 \times 10^6 + 35.7 \times 10^6)^2} \approx 0.22 \, \mathrm{m \, s^{-2}}$$ (This agrees with Table 15.1 showing approximately $0.2 \, \mathrm{m \, s^{-2}}$ for communication satellites.) **Q:** The gravitational field strength on the surface of the Moon is $1.6 \, \mathrm{N \, kg^{-1}}$. The mass of the Moon is $7.3 \times 10^{22} \, \mathrm{kg}$. What is its radius? **A:** Rearranging $g = G \frac{m}{r^2}$ to solve for $r$: $$r = \sqrt{\frac{G m}{g}}$$ $$r = \sqrt{\frac{(6.67 \times 10^{-11} \, \mathrm{N \, m^2 \, kg^{-2}}) \times (7.3 \times 10^{22} \, \mathrm{kg})}{1.6 \, \mathrm{N \, kg^{-1}}}} \approx 1.74 \times 10^6 \, \mathrm{m}$$ (The radius of the Moon is approximately 1740 km.) **Key Formulas** | Quantity | Formula | Description | |-------------------------------|--------------------------------------|------------------------------------------| | Gravitational Field Strength | $g = \frac{F_g}{m}$ | Force per unit mass | | 'g' on Planet Surface | $g = G \frac{M}{R^2}$ | For a planet of mass M, radius R | | 'g' at Altitude 'h' | $g_h = G \frac{M_E}{(R_E + h)^2}$ | $M_E$, $R_E$ for Earth, $h$ altitude | | 'g' at Altitude 'h' (simplified)| $g_h = \frac{g R_E^2}{(R_E + h)^2}$ | Using surface 'g' of Earth |