Gravitation

Step 1: Start with the formula for g at the surface of the Earth.

$$ g = \frac{GM}{R^2} \quad \text{...(i)} $$

Step 2: At a height $h$ above the surface, the distance from the center is $(R+h)$. Let the acceleration due to gravity be $g_h$.

$$ g_h = \frac{GM}{(R+h)^2} \quad \text{...(ii)} $$

Step 3: Divide equation (ii) by (i):

$$ \frac{g_h}{g} = \frac{\frac{GM}{(R+h)^2}}{\frac{GM}{R^2}} = \frac{R^2}{(R+h)^2} = \frac{R^2}{R^2(1+\frac{h}{R})^2} = \left(1 + \frac{h}{R}\right)^{-2} $$

Step 4: Apply Binomial Expansion for $h \ll R$ (neglecting higher order terms):

$$ \left(1 + \frac{h}{R}\right)^{-2} \approx 1 - \frac{2h}{R} $$

Conclusion: The value of g decreases linearly with height for small altitudes.

Step 1: Assume Earth is a uniform sphere of density $\rho$. Mass $M = \text{Volume} \times \rho$.

$$ M = \frac{4}{3}\pi R^3 \rho $$

$$ \therefore g = \frac{GM}{R^2} = \frac{G}{R^2} \left(\frac{4}{3}\pi R^3 \rho\right) = \frac{4}{3}\pi G R \rho \quad \text{...(i)} $$

Step 2: At depth $d$, the body is at a distance $(R-d)$ from the center. Only the inner sphere of radius $(R-d)$ exerts a net gravitational force. The outer shell's force cancels out.

Mass of inner sphere $M' = \frac{4}{3}\pi (R-d)^3 \rho$.

$$ g_d = \frac{GM'}{(R-d)^2} = \frac{G}{(R-d)^2} \left(\frac{4}{3}\pi (R-d)^3 \rho\right) = \frac{4}{3}\pi G (R-d) \rho \quad \text{...(ii)} $$

Step 3: Divide (ii) by (i):

$$ \frac{g_d}{g} = \frac{\frac{4}{3}\pi G (R-d) \rho}{\frac{4}{3}\pi G R \rho} = \frac{R-d}{R} = 1 - \frac{d}{R} $$

Conclusion: $g$ decreases with depth. At center ($d=R$), $g_d = 0$.

Definition: Gravitational Potential Energy at a point is the amount of work done in bringing a body of mass $m$ from infinity to that point without acceleration.

Proof:

Consider a body of mass $m$ at a distance $x$ from Earth (Mass $M$). The Gravitational Force is:

$$ F = \frac{GMm}{x^2} $$

Work done to move it by a small distance $dx$ towards the Earth (force and displacement are in same direction):

$$ dW = F \cdot dx = \frac{GMm}{x^2} dx $$

Total work done in bringing the body from infinity ($x = \infty$) to a point $P$ at distance $r$ ($x = r$) is obtained by integration:

$$ W = \int_{\infty}^{r} \frac{GMm}{x^2} dx $$

$$ W = GMm \int_{\infty}^{r} x^{-2} dx = GMm \left[ \frac{x^{-1}}{-1} \right]_{\infty}^{r} $$

$$ W = -GMm \left[ \frac{1}{x} \right]_{\infty}^{r} = -GMm \left( \frac{1}{r} - \frac{1}{\infty} \right) $$

Since this work is stored as potential energy ($U = W$):

Note: The negative sign indicates that the potential energy decreases as the body moves closer to the Earth, and the force is attractive.

Definition: The minimum velocity required to project a body vertically upwards from the surface of earth so that it escapes the gravitational field of earth.

Proof (Work-Energy Theorem):

Let a body of mass $m$ be projected with velocity $v_e$ from the surface of Earth ($R$).

1. Kinetic Energy given at surface: $K = \frac{1}{2}mv_e^2$

2. Work done by gravity as body moves from surface ($R$) to infinity ($\infty$) is negative of PE change, or we can say work done against gravity to reach infinity:

$$ W = \int_{R}^{\infty} \frac{GMm}{x^2} dx = GMm \left[ -\frac{1}{x} \right]_{R}^{\infty} $$

$$ W = -GMm \left( \frac{1}{\infty} - \frac{1}{R} \right) = \frac{GMm}{R} $$

For escape, Initial Kinetic Energy $\ge$ Work Required to escape:

$$ \frac{1}{2}mv_e^2 = \frac{GMm}{R} $$

$$ v_e^2 = \frac{2GM}{R} \Rightarrow v_e = \sqrt{\frac{2GM}{R}} $$

Since $g = \frac{GM}{R^2} \Rightarrow GM = gR^2$, we can substitute:

For Earth, $g=9.8, R=6.4\times 10^6 \Rightarrow v_e \approx 11.2 \text{ km/s}$.