Change in Gravitational Potential Energy, Energy Conservation Questions in English

(A) The gravitational potential energy of a body of mass $m$ at the surface of the earth is given by $U_1 = -\frac{G M m}{R}$.
At a height $h = \frac{R}{3}$ above the surface,the distance from the center of the earth is $r = R + h = R + \frac{R}{3} = \frac{4R}{3}$.
The gravitational potential energy at this height is $U_2 = -\frac{G M m}{r} = -\frac{G M m}{4R/3} = -\frac{3 G M m}{4R}$.
The work done in raising the body is equal to the change in gravitational potential energy:
$W = U_2 - U_1 = \left( -\frac{3 G M m}{4R} \right) - \left( -\frac{G M m}{R} \right)$.
$W = \frac{G M m}{R} - \frac{3 G M m}{4R} = \frac{4 G M m - 3 G M m}{4R} = \frac{G M m}{4R}$.

(A) The escape velocity is given by $v_E = \sqrt{\frac{2GM}{R}}$.
At the Earth's surface,the total mechanical energy is $E_1 = KE_1 + PE_1 = \frac{1}{2}mv^2 - \frac{GMm}{R}$.
Substituting $v = \alpha v_E = \alpha \sqrt{\frac{2GM}{R}}$,we get $KE_1 = \frac{1}{2}m(\alpha^2 \cdot \frac{2GM}{R}) = \frac{GMm\alpha^2}{R}$.
So,$E_1 = \frac{GMm\alpha^2}{R} - \frac{GMm}{R} = \frac{GMm}{R}(\alpha^2 - 1)$.
At the maximum height $h$,the velocity is $0$,so $KE_2 = 0$ and $PE_2 = -\frac{GMm}{R+h}$.
By the law of conservation of energy,$E_1 = E_2$:
$\frac{GMm}{R}(\alpha^2 - 1) = -\frac{GMm}{R+h}$.
$\alpha^2 - 1 = -\frac{R}{R+h} = -\frac{6400}{6400+800} = -\frac{6400}{7200} = -\frac{8}{9}$.
$\alpha^2 = 1 - \frac{8}{9} = \frac{1}{9}$.
Therefore,$\alpha = \frac{1}{3}$.

(B) The gravitational potential energy on the surface of the earth is given by $U_1 = -\frac{GMm}{R}$,where $M$ is the mass of the earth and $R$ is the radius of the earth.
At a height $h = 3R$,the distance from the center of the earth is $r = R + h = R + 3R = 4R$.
The gravitational potential energy at this height is $U_2 = -\frac{GMm}{4R}$.
The change in gravitational potential energy is $\Delta U = U_2 - U_1$.
$\Delta U = -\frac{GMm}{4R} - \left(-\frac{GMm}{R}\right) = \frac{GMm}{R} - \frac{GMm}{4R} = \frac{3GMm}{4R}$.
Using the relation $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this into the expression for $\Delta U$:
$\Delta U = \frac{3(gR^2)m}{4R} = \frac{3}{4}mgR$.

(A) Let $m = 1 \ kg$ be the mass of each sphere and $r = 2 \ m$ be the radius. The initial distance between the centers of any two spheres is $d_i = 10 \ m$.
At the time of collision,the spheres touch each other. Since they are identical,the distance between the centers of any two spheres at collision is $d_f = 2r = 2 \times 2 = 4 \ m$.
By the law of conservation of energy,the total energy at the initial position equals the total energy at the final position: $U_i + K_i = U_f + K_f$.
Initially,$K_i = 0$. The potential energy of the system of three spheres is $U = -3 \times \frac{G m^2}{d}$.
So,$-3 \frac{G m^2}{d_i} = -3 \frac{G m^2}{d_f} + 3 \times (\frac{1}{2} m v^2)$,where $v$ is the speed of each sphere.
Dividing by $3m/2$,we get $v^2 = 2 G m (\frac{1}{d_f} - \frac{1}{d_i})$.
Substituting the values: $v^2 = 2 \times G \times 1 \times (\frac{1}{4} - \frac{1}{10}) = 2 G (\frac{5-2}{20}) = 2 G (\frac{3}{20}) = \frac{3 G}{10}$.
Therefore,$v = \sqrt{\frac{3 G}{10}}$.

(D) The gravitational potential energy of a body of mass $m$ at a distance $r$ from the center of the earth is given by $U = -\frac{GMm}{r}$.
At the surface of the earth,$r = R$,so $U_i = -\frac{GMm}{R}$.
For a height $h = R$,the distance from the center is $r = R + R = 2R$. The potential energy is $U_f = -\frac{GMm}{2R}$.
The energy required $W$ is the change in potential energy:
$W = U_f - U_i = -\frac{GMm}{2R} - (-\frac{GMm}{R}) = \frac{GMm}{2R}$.
For a height $h = 2R$,the distance from the center is $r = R + 2R = 3R$. The potential energy is $U_f' = -\frac{GMm}{3R}$.
The energy required $W'$ is:
$W' = U_f' - U_i = -\frac{GMm}{3R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{3R} = \frac{2GMm}{3R}$.
Since $W = \frac{GMm}{2R}$,we have $\frac{GMm}{R} = 2W$.
Substituting this into the expression for $W'$:
$W' = \frac{2}{3} \times (2W) = \frac{4W}{3}$.

(C) Using the law of conservation of energy, the total energy at the surface equals the total energy at the maximum height $h$.
At the surface: $E_i = K_i + U_i = \frac{1}{2}mv^2 - \frac{GMm}{R}$
At maximum height: $E_f = K_f + U_f = 0 - \frac{GMm}{R+h}$
Equating $E_i = E_f$: $\frac{1}{2}mv^2 - \frac{GMm}{R} = - \frac{GMm}{R+h}$
Dividing by $m$ and using $GM = gR^2$: $\frac{v^2}{2} - gR = - \frac{gR^2}{R+h}$
Rearranging: $\frac{gR^2}{R+h} = gR - \frac{v^2}{2} = gR(1 - \frac{v^2}{2gR})$
$\frac{R}{R+h} = 1 - \frac{v^2}{2gR} = 1 - \frac{(8000)^2}{2 \times 10 \times 6400 \times 10^3} = 1 - \frac{64 \times 10^6}{128 \times 10^6} = 1 - 0.5 = 0.5$
$\frac{R}{R+h} = 0.5 \implies R+h = 2R \implies h = R$
Since $R = 6400 \,km$, the maximum height $h = 6400 \,km$.

(A) According to the law of conservation of mechanical energy,the total energy at the surface of the earth must equal the total energy at the maximum height $h$.
At the surface: $E_i = K_i + U_i = \frac{1}{2} m v_0^2 - \frac{G M m}{R}$.
At maximum height $h$: $E_f = K_f + U_f = 0 - \frac{G M m}{R+h}$.
Equating $E_i = E_f$:
$\frac{1}{2} m v_0^2 - \frac{G M m}{R} = - \frac{G M m}{R+h}$.
Dividing by $m$ and rearranging:
$\frac{v_0^2}{2} = G M \left( \frac{1}{R} - \frac{1}{R+h} \right) = G M \left( \frac{h}{R(R+h)} \right)$.
Since $g = \frac{G M}{R^2}$,we have $G M = g R^2$.
Substituting $G M$:
$\frac{v_0^2}{2} = g R^2 \left( \frac{h}{R(R+h)} \right) = \frac{g R h}{R+h}$.
$v_0^2 (R+h) = 2 g R h$.
$v_0^2 R + v_0^2 h = 2 g R h$.
$v_0^2 R = h (2 g R - v_0^2)$.
$h = \frac{R v_0^2}{2 g R - v_0^2}$.

(B) The gravitational potential energy of an object of mass $m$ at a distance $r$ from the center of the earth is given by $U = -\frac{GMm}{r}$.
At the surface of the earth,$r = R$,so $U_i = -\frac{GMm}{R}$.
At a height $h = R$ above the surface,the distance from the center is $r = R + h = R + R = 2R$.
Thus,the potential energy at height $h$ is $U_f = -\frac{GMm}{2R}$.
The gain in potential energy is $\Delta U = U_f - U_i = -\frac{GMm}{2R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}$.
Since the acceleration due to gravity on the surface is $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this into the expression for $\Delta U$,we get $\Delta U = \frac{(gR^2)m}{2R} = \frac{mgR}{2}$.

(A) The gravitational potential energy $U$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.
At the surface of the Earth,$r = R$,so $U_i = -\frac{GMm}{R}$.
At a height $h = nR$ above the surface,the distance from the center is $r = R + nR = R(1+n)$.
So,$U_f = -\frac{GMm}{R(1+n)}$.
The change in gravitational potential energy is $\Delta U = U_f - U_i = -\frac{GMm}{R(1+n)} - (-\frac{GMm}{R})$.
$\Delta U = \frac{GMm}{R} \left(1 - \frac{1}{1+n}\right) = \frac{GMm}{R} \left(\frac{1+n-1}{1+n}\right) = \frac{GMm}{R} \left(\frac{n}{1+n}\right)$.
Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this,$\Delta U = \frac{gR^2 m}{R} \left(\frac{n}{n+1}\right) = \left(\frac{n}{n+1}\right) mgR$.

(A) The work done $W$ in moving a mass $m$ from the surface of the Earth to a height $h$ is given by the change in gravitational potential energy: $W = U_f - U_i$.
Here,the initial potential energy at the surface is $U_i = -\frac{GMm}{R}$.
The final potential energy at height $h = R$ is $U_f = -\frac{GMm}{R+R} = -\frac{GMm}{2R}$.
Therefore,$W = -\frac{GMm}{2R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}$.
Since the acceleration due to gravity at the surface is $g = \frac{GM}{R^2}$,we can write $GM = gR^2$.
Substituting this into the expression for work: $W = \frac{(gR^2)m}{2R} = \frac{mgR}{2}$.