Gravitation

(a) Gravitational Shielding:
NO. Unlike electric forces, gravitational force is independent of the intervening medium. You cannot shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere.

(b) Detecting gravity in space station:
YES. If the space station is large enough, the astronaut can detect gravity due to tidal forces (difference in gravitational force) or if the station rotates (artificial gravity). However, strictly in free fall, the local frame is inertial (weightless). For a very large station, the variation in $g$ from center to edge might be detectable.

(c) Tides (Moon vs Sun):
[Image of earth tides caused by moon vs sun]
Tidal effect depends on the inverse cube of distance ($1/r^3$), while gravitational force depends on inverse square ($1/r^2$). Since the Moon is much closer, the differential pull (tidal force) across Earth’s diameter is significantly larger for the Moon than for the Sun, even though the Sun’s total pull is stronger.

(a) Altitude: Acceleration due to gravity DECREASES with increasing altitude ($g \propto 1/r^2$).

(b) Depth: Acceleration due to gravity DECREASES with increasing depth ($g_{depth} = g(1-d/R)$). At center, $g=0$.

(c) Independence: Acceleration due to gravity is independent of the MASS OF THE BODY (falls at same rate).

(d) Potential Energy Formula: The formula $-GMm(1/r_2 – 1/r_1)$ is MORE ACCURATE than $mg(r_2-r_1)$. The linear formula assumes $g$ is constant, which is only true for small distances near the surface.

Given: $T_P = \frac{1}{2} T_E$.
According to Kepler’s Third Law: $T^2 \propto R^3$.

$\frac{R_P}{R_E} = \left(\frac{1}{2}\right)^{2/3} = (0.25)^{1/3} \approx \mathbf{0.63}$.
The orbital size is 0.63 times that of Earth.

Formula: $M = \frac{4\pi^2 R^3}{G T^2}$.

• Sun-Earth: $M_S = \frac{4\pi^2 R_E^3}{G T_E^2}$. ($R_E = 1.5 \times 10^{11}$ m, $T_E = 365$ days).
• Jupiter-Io: $M_J = \frac{4\pi^2 R_{Io}^3}{G T_{Io}^2}$. ($R_{Io} = 4.22 \times 10^8$ m, $T_{Io} = 1.769$ days).

Ratio $\frac{M_S}{M_J} = \left(\frac{R_E}{R_{Io}}\right)^3 \left(\frac{T_{Io}}{T_E}\right)^2$.
$\frac{M_S}{M_J} \approx (355)^3 \times (0.0048)^2 \approx \mathbf{1000}$.
Thus, Mass of Jupiter $\approx 1/1000$ Mass of Sun.

Given: $r = 50,000 \text{ ly}$. Total Mass $M = 2.5 \times 10^{11} M_{Sun}$.
We assume the star orbits the center of mass of the galaxy. Using Kepler’s Law form:

Convert units to SI or simply scaling. Let’s use SI.
$1 \text{ ly} = 9.46 \times 10^{15} \text{ m}$. $M_{Sun} = 2 \times 10^{30} \text{ kg}$.
Calculating… $T \approx \mathbf{3.5 \times 10^8 \text{ years}}$.

(a) Energy of Orbiting Satellite:
$K = \frac{GMm}{2r}$, $U = -\frac{GMm}{r}$. Total $E = -\frac{GMm}{2r}$.
Total energy is negative of its KINETIC energy.

(b) Launch Energy:
Stationary object: Needs energy to reach infinity from height $h$. $E_{req} = 0 – (-GMm/r) = GMm/r$.
Orbiting satellite: Already has Kinetic Energy. $E_{req} = 0 – (-GMm/2r) = GMm/2r$.
Energy required for orbiting satellite is LESS.

$v_e = \sqrt{2GM/R}$.

• (a) Mass of body: NO.
• (b) Location of projection: YES (depends on latitude/radius slightly due to Earth’s shape).
• (c) Direction of projection: NO.
• (d) Height of location: YES ($v_e = \sqrt{2GM/(R+h)}$).

(a) Swollen Feet: No. Gravity usually pulls blood to feet; lack of gravity stops this.
(b) Swollen Face: YES. In microgravity, fluids redistribute towards the upper body/head (“Puffy Face Bird Leg” syndrome).
(c) Headache: YES. Due to fluid shift and increased intracranial pressure.
(d) Orientational Problem: YES. Inner ear relies on gravity for orientation.

Let $x$ be distance from Earth. Distance from Sun is $r – x$.
Force equality: $\frac{G M_e m}{x^2} = \frac{G M_s m}{(r-x)^2}$.

$r – x = 577.35 x \Rightarrow 578.35 x = r$.
$x = \frac{1.5 \times 10^{11}}{578.35} \approx \mathbf{2.59 \times 10^8 \text{ m}}$.

Weight on surface $W = mg = 63 \text{ N}$.
At height $h = R/2$, distance from center $r = R + R/2 = 3R/2$.
Gravity $g’ = \frac{GM}{r^2} = \frac{GM}{(1.5R)^2} = \frac{g}{2.25}$.

New Weight $W’ = m g’ = \frac{mg}{2.25} = \frac{63}{2.25} = \mathbf{28 \text{ N}}$.

Weight on surface $W = 250 \text{ N}$. Depth $d = R/2$.
Formula: $g’ = g(1 – d/R)$.

New Weight $W’ = 0.5 W = \mathbf{125 \text{ N}}$.

Escape velocity $v_e = 11.2 \text{ km/s}$. Projection speed $v_0 = 3v_e$.
Conservation of Energy:
$\frac{1}{2}m v_0^2 – \frac{GMm}{R} = \frac{1}{2}m v_f^2 + 0$ (at infinity).

$\frac{1}{2}m(3v_e)^2 – \frac{1}{2}m v_e^2 = \frac{1}{2}m v_f^2$.
$9v_e^2 – v_e^2 = v_f^2 \Rightarrow v_f = \sqrt{8} v_e$.
$v_f = 2.828 \times 11.2 \approx \mathbf{31.7 \text{ km/s}}$.

Satellite mass $m = 200 \text{ kg}$. Height $h = 400 \text{ km}$.
Orbit radius $r = 6400 + 400 = 6800 \text{ km} = 6.8 \times 10^6 \text{ m}$.

Total Energy in orbit $E_{orbit} = -\frac{GMm}{2r}$.
Energy at infinity $E_{\infty} = 0$.
Energy required $\Delta E = E_{\infty} – E_{orbit} = \frac{GMm}{2r}$.

$\Delta E = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 200}{2 \times 6.8 \times 10^6}$.
$\Delta E \approx \mathbf{5.9 \times 10^9 \text{ J}}$.

Initial distance $r = 10^9 \text{ km}$. Initial speed $\approx 0$.
Final distance (collision) $2R = 2 \times 10^4 \text{ km}$.
Conservation of Energy:
$K_i + U_i = K_f + U_f$.
$0 – \frac{GMM}{r} = 2 \times (\frac{1}{2}Mv^2) – \frac{GMM}{2R}$.
$v^2 = GM(\frac{1}{2R} – \frac{1}{r})$. Since $r \gg 2R$, ignore $1/r$.
$v = \sqrt{\frac{GM}{2R}} = \sqrt{\frac{6.67 \times 10^{-11} \times 2 \times 10^{30}}{2 \times 10^7}}$.
$v \approx \mathbf{2.6 \times 10^6 \text{ m/s}}$.

Two spheres $M = 100 \text{ kg}$, separation $r = 1.0 \text{ m}$. Midpoint $d = 0.5 \text{ m}$.

Force: The pulls from both spheres are equal and opposite.
Net Force $F = \frac{GMm}{d^2} – \frac{GMm}{d^2} = \mathbf{0}$.
Equilibrium: Since $F=0$, the object is in equilibrium.

Potential: Potentials add up (scalars).
$V = -\frac{GM}{d} – \frac{GM}{d} = -\frac{2GM}{d}$.
$V = -\frac{2 \times 6.67 \times 10^{-11} \times 100}{0.5} = \mathbf{-2.7 \times 10^{-8} \text{ J/kg}}$.

Stability: If moved slightly towards one sphere, the force from that sphere increases (closer distance), pulling it further away from the center. Thus, the equilibrium is UNSTABLE.