Step 1: Given Variables and Initial Setup

To determine if an object remains stuck to the surface, we must compare the gravitational acceleration ($g'$) at the equator with the centripetal acceleration ($a_c$) required for circular motion due to the rotation of the star.

• Mass of the star ($M$): $2.5 \times M_{sun} = 2.5 \times (2 \times 10^{30} \text{ kg}) = 5.0 \times 10^{30} \text{ kg}$
• Radius of the star ($R$): $12 \text{ km} = 1.2 \times 10^4 \text{ m}$
• Frequency of rotation ($f$): $1.2 \text{ rev/s}$
• Gravitational constant ($G$): $6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$

Step 2: Calculating Gravitational Acceleration ($g'$)

The acceleration due to gravity at the surface of the star is given by the formula:

$g' = \frac{GM}{R^2}$

Substituting the values:

$g' = \frac{(6.67 \times 10^{-11}) \times (5.0 \times 10^{30})}{(1.2 \times 10^4)^2}$

$g' = \frac{33.35 \times 10^{19}}{1.44 \times 10^8} \approx 2.316 \times 10^{12} \text{ m/s}^2$ [Per Newton's Law of Universal Gravitation]

Step 3: Calculating Centripetal Acceleration ($a_c$)

The centripetal acceleration required to maintain an object on the surface is $a_c = \omega^2 R$, where $\omega = 2\pi f$.

$\omega = 2 \times 3.14159 \times 1.2 \approx 7.54 \text{ rad/s}$

$a_c = (7.54)^2 \times (1.2 \times 10^4)$

$a_c \approx 56.85 \times 1.2 \times 10^4 \approx 6.82 \times 10^5 \text{ m/s}^2$

Step 4: Comparison and Conclusion

An object will remain stuck to the surface if the gravitational pull exceeds the required centripetal acceleration ($g' > a_c$).

Parameter	Magnitude ($\text{m/s}^2$)
Gravitational Acceleration ($g'$)	$2.316 \times 10^{12}$
Centripetal Acceleration ($a_c$)	$6.82 \times 10^5$

Since $g' \gg a_c$ (by several orders of magnitude), the gravitational force is significantly stronger than the centrifugal effect of the rotation.

Final Solution: Yes, the object will remain stuck to the surface because the gravitational acceleration ($2.316 \times 10^{12} \text{ m/s}^2$) is vastly greater than the acceleration required to fling it off the surface ($6.82 \times 10^5 \text{ m/s}^2$).

Given Variables & Initial Setup

To launch a spaceship out of the solar system, we must provide sufficient energy to overcome the gravitational potential energy exerted by both the Sun and Mars. The total energy required ($E$) is the negative of the total mechanical energy of the spaceship in its current position.

• Mass of spaceship ($m$) = $1000 \, \text{kg}$
• Mass of Sun ($M_S$) = $2 \times 10^{30} \, \text{kg}$
• Mass of Mars ($M_M$) = $6.4 \times 10^{23} \, \text{kg}$
• Radius of Mars ($R_M$) = $3395 \times 10^3 \, \text{m} = 3.395 \times 10^6 \, \text{m}$
• Radius of Mars' orbit ($r$) = $2.28 \times 10^8 \, \text{km} = 2.28 \times 10^{11} \, \text{m}$
• Gravitational constant ($G$) = $6.67 \times 10^{-11} \, \text{N m}^2 \text{kg}^{-2}$

Step 1: Calculating the Potential Energy at the Surface of Mars

The total energy ($E_{total}$) is the sum of the potential energy due to the Sun ($U_S$) and the potential energy due to Mars ($U_M$), plus the kinetic energy ($K$) of the spaceship due to Mars' orbital motion. [Per the Law of Conservation of Energy, $E_{launch} = -E_{total}$].

Potential energy due to Mars ($U_M$):

$U_M = -\frac{GM_Mm}{R_M}$ $U_M = -\frac{(6.67 \times 10^{-11})(6.4 \times 10^{23})(1000)}{3.395 \times 10^6}$ $U_M \approx -1.258 \times 10^{10} \, \text{J}$

Step 2: Calculating Potential Energy due to the Sun

The spaceship is at the distance of Mars from the Sun.

$U_S = -\frac{GM_Sm}{r}$ $U_S = -\frac{(6.67 \times 10^{-11})(2 \times 10^{30})(1000)}{2.28 \times 10^{11}}$ $U_S \approx -5.851 \times 10^{11} \, \text{J}$

Step 3: Accounting for Kinetic Energy of the Spaceship

The spaceship possesses kinetic energy due to the orbital velocity ($v_o$) of Mars. [Per the principle of orbital mechanics, $v_o = \sqrt{\frac{GM_S}{r}}$].

$K = \frac{1}{2}mv_o^2 = \frac{1}{2}m\left(\frac{GM_S}{r}\right)$ $K = \frac{(1000)(6.67 \times 10^{-11})(2 \times 10^{30})}{2 \times 2.28 \times 10^{11}}$ $K \approx 2.925 \times 10^{11} \, \text{J}$

Step 4: Total Energy Required

The total mechanical energy $E_{total} = U_M + U_S + K$. The energy required to escape ($E_{expended}$) is $-E_{total}$.

$E_{total} = (-0.1258 \times 10^{11}) + (-5.851 \times 10^{11}) + (2.925 \times 10^{11})$ $E_{total} = -2.9268 \times 10^{11} \, \text{J}$

Energy to be expended:

$E_{expended} = -(-2.9268 \times 10^{11} \, \text{J})$ $E_{expended} \approx 2.93 \times 10^{11} \, \text{J}$

Visualization of Gravitational Well

Final Solution: The energy required to launch the spaceship out of the solar system is approximately $2.93 \times 10^{11} \, \text{J}$.

Step 1: Given Variables and Constants

• Initial velocity of the rocket, $v = 2 \times 10^3 \, \text{m s}^{-1}$
• Mass of Mars, $M = 6.4 \times 10^{23} \, \text{kg}$
• Radius of Mars, $R = 3395 \times 10^3 \, \text{m}$
• Gravitational Constant, $G = 6.67 \times 10^{-11} \, \text{N m}^2 \text{kg}^{-2}$
• Energy loss factor, $\eta = 20\% = 0.2$

Step 2: Energy Conservation Analysis

The total initial mechanical energy ($E_i$) of the rocket at the surface consists of kinetic energy ($K$) and potential energy ($U$). [Per the Principle of Conservation of Energy].

$E_i = K + U = \frac{1}{2}mv^2 - \frac{GMm}{R}$

Since 20% of the initial kinetic energy is lost to atmospheric resistance, the effective energy available at the surface to reach the maximum height $h$ is:

$E_{effective} = (K - 0.2K) + U = 0.8 \left( \frac{1}{2}mv^2 \right) - \frac{GMm}{R}$

$E_{effective} = 0.4mv^2 - \frac{GMm}{R}$

Step 3: Energy at Maximum Height

At the maximum height ($R+h$), the velocity of the rocket is zero. Therefore, the total energy ($E_f$) at that point is purely potential:

$E_f = -\frac{GMm}{R+h}$

Equating $E_{effective}$ to $E_f$ [By the Work-Energy Theorem]:

$0.4mv^2 - \frac{GMm}{R} = -\frac{GMm}{R+h}$

Dividing by mass $m$ (as it cancels out):

$0.4v^2 - \frac{GM}{R} = -\frac{GM}{R+h}$

Step 4: Numerical Computation

Calculating the individual components:

• $0.4v^2 = 0.4 \times (2000)^2 = 0.4 \times 4 \times 10^6 = 1.6 \times 10^6 \, \text{J kg}^{-1}$
• $\frac{GM}{R} = \frac{6.67 \times 10^{-11} \times 6.4 \times 10^{23}}{3395 \times 10^3} \approx 12.58 \times 10^6 \, \text{J kg}^{-1}$

Rearranging for $R+h$:

$\frac{GM}{R+h} = \frac{GM}{R} - 0.4v^2$

$\frac{GM}{R+h} = 12.58 \times 10^6 - 1.6 \times 10^6 = 10.98 \times 10^6$

$R+h = \frac{6.67 \times 10^{-11} \times 6.4 \times 10^{23}}{10.98 \times 10^6} \approx 3.887 \times 10^6 \, \text{m}$

Step 5: Final Displacement

$h = (R+h) - R = 3887 \, \text{km} - 3395 \, \text{km} = 492 \, \text{km}$

Final Solution: The rocket will reach a maximum distance of 492 km from the surface of Mars.

Step 1: Identification of Given Variables

To determine the gravitational potential $V$ at the site of a geostationary satellite, we define the parameters provided:

• Mass of Earth ($M$): $6.0 \times 10^{24}\ \text{kg}$
• Radius of Earth ($R$): $6400\ \text{km} = 6.4 \times 10^6\ \text{m}$
• Height of satellite ($h$): $36,000\ \text{km} = 3.6 \times 10^7\ \text{m}$
• Gravitational constant ($G$): $6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$

Step 2: Defining the Orbital Radius

The gravitational potential is dependent on the distance from the center of the Earth. The distance $r$ from the center to the satellite is the sum of Earth's radius and the altitude:

$r = R + h$

$r = 6.4 \times 10^6\ \text{m} + 36 \times 10^6\ \text{m}$

$r = 4.24 \times 10^7\ \text{m}$ [By definition of radial distance in a central force field]

Step 3: Formulating the Potential Equation

The gravitational potential $V$ at a point at distance $r$ from a point mass (or spherical mass) $M$ is defined as the work done per unit mass to bring a test mass from infinity to that point [Per the definition of Gravitational Potential Energy $U = -GMm/r$]:

$V = -\frac{GM}{r}$

Step 4: Calculation

Substituting the values into the derived formula:

$V = -\frac{(6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2) \times (6.0 \times 10^{24}\ \text{kg})}{4.24 \times 10^7\ \text{m}}$

$V = -\frac{40.02 \times 10^{13}}{4.24 \times 10^7}\ \text{J/kg}$

$V \approx -9.438 \times 10^6\ \text{J/kg}$

Step 5: Analysis of Results

The negative sign indicates that the gravitational force is attractive and that the potential is lower than the potential at infinity (which is defined as zero) [Per the convention of gravitational field theory].

Final Solution: The gravitational potential at the site of the geostationary satellite is approximately $-9.44 \times 10^6\ \text{J/kg}$.