What is gravity at height h above surface?

$g_h = g\left(\frac{R}{R+h}\right)^2$. At height $h = R/3$: $g_h = g\left(\frac{R}{R+R/3}\right)^2 = g\left(\frac{3}{4}\right)^2 = \frac{9g}{16}$.

What is the new weight?

Weight $= mg_h = 48 \times \frac{9}{16} = \frac{432}{16} = 27$ N.

Why does weight decrease with altitude?

Gravitational force $\propto 1/r^2$. As $r$ increases (height above surface), gravitational force decreases. At $r = R+h$: $g_h = GM/(R+h)^2 = g\cdot R^2/(R+h)^2$.

What is weight at depth d inside earth?

$g_d = g(1 - d/R)$. At centre ($d=R$): $g = 0$. Weight decreases linearly with depth (unlike altitude where it decreases as $1/r^2$).

At what height does weight become half?

$g_h = g/2$: $R^2/(R+h)^2 = 1/2$, so $R+h = R\sqrt{2}$, $h = R(\sqrt{2}-1) \approx 0.414R$.

The weight of a body on the surface of earth is 48 N. What is its weight at a height equal to $R/3$ from the surface of

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The weight of a body on the surface of earth is 48 N. What is its weight at a height equal to $R/3$ from the surface of earth? (R = radius of earth)

Options

$27$ N

$36$ N

$32$ N

$24$ N

Correct Answer

27 N

Solution

$$g_h = g\left(\frac{R}{R+h}\right)^2, \quad h = \frac{R}{3}$$$$g_h = g\left(\frac{R}{R+R/3}\right)^2 = g\left(\frac{R}{4R/3}\right)^2 = g\left(\frac{3}{4}\right)^2 = \frac{9g}{16}$$

$$W_h = mg_h = 48 \times \frac{9}{16} = \boxed{27 \text{ N}}$$

$g_h = g\left(\dfrac{R}{R+h}\right)^2 = \dfrac{9g}{16}$ at $h=R/3$
Weight $= 48\times 9/16 = 27$ N

Theory: Gravitation

1. Variation of g with Altitude

$g_h = \frac{GM}{(R+h)^2} = g\left(\frac{R}{R+h}\right)^2$. For small heights ($h \ll R$): $g_h \approx g(1 - 2h/R)$. Weight $= mg_h$ decreases with altitude. At $h=R$: $g_h = g/4$. At $h=2R$: $g_h = g/9$. At $h = (\sqrt{2}-1)R \approx 0.414R$: $g_h = g/2$. There is no point in space (other than infinity) where gravity is exactly zero — it only approaches zero asymptotically.

2. Variation of g with Depth

$g_d = g\left(1 - \frac{d}{R}\right)$ (for uniform density Earth). Decreases linearly with depth. At centre ($d=R$): $g=0$. Decreases as you go below surface. This is different from altitude (where it decreases as $1/r^2$). Maximum $g$ is at Earth surface. $g$ decreases both above AND below the surface. The two expressions (altitude and depth) give the same result only at surface ($h=0$ or $d=0$).

3. Variation of g with Latitude

$g_{eff} = g - \omega^2 R\cos^2\phi$ where $\phi$ = latitude, $\omega$ = Earth rotation speed. At equator ($\phi=0$): $g_{eff} = g - \omega^2 R$ (minimum). At poles ($\phi=90°$): $g_{eff} = g$ (maximum, no centrifugal effect). Also: Earth is oblate (flatter at poles) → poles are closer to Earth centre → stronger gravity. $g_{poles} - g_{equator} \approx 0.052$ m/s². $g$ also varies with local geology (ore deposits affect local $g$).

4. Gravitational Potential Energy

$U = -\frac{GMm}{r}$ (zero at infinity). At Earth surface: $U = -\frac{GMm}{R} = -mgR$. To escape to infinity: need $\frac{1}{2}mv_e^2 = mgR$, so $v_e = \sqrt{2gR} = \sqrt{2GM/R}$. $v_e = 11.2$ km/s for Earth. On Moon ($g = 1.6$ m/s², $R = 1.74\times10^6$ m): $v_e = 2.38$ km/s (easy for gas molecules to escape → no atmosphere on Moon).

5. Satellites and Orbital Mechanics

For circular orbit at height $h$: $v_o = \sqrt{g_h(R+h)} = \sqrt{GM/(R+h)}$. Period: $T = 2\pi(R+h)/v_o = 2\pi\sqrt{(R+h)^3/(GM)}$ (Kepler III). Geostationary orbit: $T=24$ h, $h \approx 36,000$ km. Low Earth orbit (LEO): $h = 400$ km, $T \approx 90$ min. Orbital energy: $E = -GMm/(2r) = -mgR^2/(2r)$. Negative energy means bound orbit.

6. Tides

Tides: differential gravitational force of Moon on Earth. Moon pulls near side more strongly than far side. Creates two tidal bulges: one toward Moon, one away (inertia). Two high tides per day. Spring tides (Sun, Moon, Earth aligned): stronger tides. Neap tides (right angle): weaker. Tidal friction: Earth rotation slowing by 1.4 ms/century. Angular momentum transferred to Moon → Moon receding at 3.8 cm/year. Eventually: tidal locking of Earth to Moon (day = month ≈ 47 current days).

7. Gravitational Field and Gauss Law

Gravitational field $\vec{g} = -GM\hat{r}/r^2$ (toward mass). Gravitational flux: $\oint \vec{g}\cdot d\vec{A} = -4\pi G M_{enc}$. Analogous to Gauss law but with $-4\pi G$ instead of $1/\epsilon_0$. Inside uniform sphere: $g \propto r$ (increases linearly from 0 at centre). Outside: $g \propto 1/r^2$. Gravitational potential: $V = -GM/r$. $\vec{g} = -\nabla V$. Equipotential surfaces: concentric spheres around a point mass.

8. Gravitational Waves

Predicted by Einstein (General Relativity, 1916). Ripples in spacetime caused by accelerating massive objects. First detected by LIGO (2015) from merger of two black holes 1.3 billion light-years away. Properties: travel at speed of light. Extremely weak (LIGO detected changes of $10^{-18}$ m = 1/1000 of proton size!). Carry energy and angular momentum from inspiraling binary systems. Examples detected: binary black hole mergers, binary neutron star merger (GW170817, also detected in light → multi-messenger astronomy). Nobel Prize 2017: Weiss, Barish, Thorne.

Frequently Asked Questions

1. Why does weight decrease at height R/3? ⌄

Gravitational force $F = GMm/r^2$ where $r$ = distance from Earth centre. At surface: $r = R$, weight $= mg = 48$ N. At height $h = R/3$: $r = R + R/3 = 4R/3$. New weight $= GMm/(4R/3)^2 = GMm \times 9/(16R^2) = (9/16)\times mg = (9/16)\times 48 = 27$ N. The $r$ increased by factor $4/3$, so $r^2$ increased by $(4/3)^2 = 16/9$, so force decreased by factor $9/16$.

2. How is the formula g_h = g(R/(R+h))^2 derived? ⌄

At surface: $g = GM/R^2$. At height $h$: $g_h = GM/(R+h)^2$. Dividing: $g_h/g = R^2/(R+h)^2 = (R/(R+h))^2$. Therefore $g_h = g(R/(R+h))^2$. This exact formula works for any height (no approximation). The approximation $g_h \approx g(1-2h/R)$ (first-order Taylor expansion) is only valid for $h \ll R$.

3. At what height does weight become 1/4 of surface weight? ⌄

$W_h = W/4$: $g_h = g/4$. $(R/(R+h))^2 = 1/4$. $R/(R+h) = 1/2$. $2R = R+h$. $h = R$. At height equal to Earth radius ($h = R \approx 6400$ km), weight becomes 1/4. At $h = 2R$: weight becomes 1/9. At $h = nR$: weight becomes $1/(n+1)^2$ of surface weight. Pattern: $W_h/W = 1/(1+h/R)^2$.

4. Does mass change with altitude? ⌄

No. Mass is an intrinsic property of matter — it does not change with location, altitude, or gravitational field. Only weight (force of gravity = $mg$) changes. On Moon: mass of 60 kg person = 60 kg (same as Earth). Weight on Moon = $60 \times 1.6 = 96$ N (compared to $60 \times 9.8 = 588$ N on Earth). This is why we distinguish mass (kg) from weight (N or kgf). Astronauts in orbit: mass unchanged, apparent weight = 0 (weightlessness due to free fall).

5. How does g vary inside Earth? ⌄

Inside Earth (assuming uniform density): $g_d = g(1 - d/R)$ or equivalently $g_r = g\cdot r/R$ where $r$ = distance from centre and $d = R-r$ = depth. This is because only the mass in the sphere of radius $r$ contributes (shell theorem: mass outside $r$ creates no net force). At $r = 0$ (centre): $g = 0$. At $r = R$ (surface): $g$ = normal surface value. Actual Earth: non-uniform density (denser core) → $g$ actually INCREASES slightly going into crust, reaches maximum at core-mantle boundary, then decreases to 0 at centre.

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