Class 11 Physics (Physics Part-I) Chapter 7 Gravitation

Key Concept: Comparative Analysis

b) Potential energy is lower at $1000 \, km$ than at $2000 \, km$

[Solution Description]

The potential energy ($PE$) of a satellite in orbit is calculated using the formula $PE = -\frac{GMm}{R_E + h}$. We’ll find $PE$ for each altitude.

For $h_1 = 1000 \, km$:

$PE_1 = -\frac{6.674 \times 10^{-11} \times 5.97 \times 10^{24} \times 1000}{(6371 + 1000) \times 10^3}$

$PE_1 \approx -5.874 \times 10^{10} \, J$

For $h_2 = 2000 \, km$:

$PE_2 = -\frac{6.674 \times 10^{-11} \times 5.97 \times 10^{24} \times 1000}{(6371 + 2000) \times 10^3}$

$PE_2 \approx -5.270 \times 10^{10} \, J$

Comparison shows $PE_2 > PE_1$, meaning the potential energy increases (becomes less negative) as altitude increases.

Key Concept: Comparative Analysis

d) Jupiter

[Solution Description] According to Kepler’s third law, $T^2 \propto a^3$. For Earth, $a = 1$, so $T_E^2 = k \times 1^3 = k$. For Jupiter, $a = 5.2$, so $T_J^2 = k \times 5.2^3$. Comparing the two: $T_J > T_E$.

Key Concept: Height Dependency

c) It decreases by a factor of $\frac{1}{2}$

[Solution Description] Given that initial height is $h$, with orbital speed $V_1 = \sqrt{\frac{GM_E}{R_E + h}}$. When height becomes $4h$, the new orbital speed $V_2$ is calculated as follows:

$V_2 = \sqrt{\frac{GM_E}{R_E + 4h}}$

To find the ratio of speeds:

$\frac{V_2}{V_1} = \frac{\sqrt{\frac{GM_E}{R_E + 4h}}}{\sqrt{\frac{GM_E}{R_E + h}}}$

Simplifying,

$\frac{V_2}{V_1} = \sqrt{\frac{R_E + h}{R_E + 4h}}$

Since $\frac{R_E + h}{R_E + 4h}$ is less than 1, $V_2$ is less than $V_1$.

Let’s assume $R_E >> h$, then it can be approximated as:

$\frac{V_2}{V_1} \approx \frac{1}{2}$, indicating that $V_2$ is roughly half of $V_1$.

Key Concept: Energy Change

b) Decreases by $\frac{G M m}{12 R_E}$

[Solution Description] The total mechanical energy $E$ of a satellite is given by:

$E = K + V$

where $V = -\frac{G M m}{r}$ and $K = -\frac{V}{2}$.

For the initial orbit:

$E_1 = -\frac{G M m}{2 r_1}$

For the new orbit:

$E_2 = -\frac{G M m}{2 r_2}$

Substituting $r_1 = 3R_E$ and $r_2 = 6R_E$:

$E_1 = -\frac{G M m}{6 R_E}, \quad E_2 = -\frac{G M m}{12 R_E}$

The change in energy $\Delta E$ is:

$\Delta E = E_2 – E_1 = \left(-\frac{G M m}{12 R_E}\right) – \left(-\frac{G M m}{6 R_E}\right)$

$\Delta E = \frac{G M m}{12 R_E} – \frac{G M m}{6 R_E} = -\frac{G M m}{12 R_E}$

Therefore, the total mechanical energy decreases by $\frac{G M m}{12 R_E}$.

Key Concept: Kepler’s Period Law

b) Semi-major axis

[Solution Description]

Kepler’s third law states that the square of the orbital period $(T^2)$ is directly proportional to the cube of the semi-major axis $(a^3)$ of its orbit. Mathematically, this can be expressed as $T^2 \propto a^3$. This relation holds true for both circular and elliptical orbits.

Key Concept: Energy Calculation

b) Decreases

[Solution Description]

The total energy $E$ of a satellite in orbit is given by:
$E = -\frac{G m M_E}{2 (R_E + h)}$
When height is doubled:
$E_{new} = -\frac{G m M_E}{2 (R_E + 2h)}$
Comparing the energies:
$E_{new} = \frac{(R_E + h)}{(R_E + 2h)}E$
Since $R_E + 2h > R_E + h$, $E_{new}$ is greater than the initial $E$. Thus, the energy magnitude decreases.

Key Concept: Energy Concepts

c) KE is half of the absolute value of PE

[Solution Description] In a stable circular orbit, the total energy ($E_t$) of the satellite is given by $E_t = KE + PE = -\frac{GMm}{2r}$. Here, the kinetic energy ($KE$) is half the magnitude of the potential energy ($PE$), but positive: $KE = -\frac{1}{2}PE$. Thus, $KE = -\frac{1}{2}(-\frac{GMm}{r}) = \frac{GMm}{2r}$ and $PE = -\frac{GMm}{r}$.

Key Concept: Speed Relation

b) Decreases

[Solution Description] According to the equations of motion for satellites, as the height $h$ of a satellite increases, its speed decreases. This is because the gravitational pull weakens with increased distance from Earth, requiring less speed to maintain orbit.

Key Concept: Basic Concept

d) Escape speed decreases by a factor of $\sqrt{2}$

[Solution Description] The escape speed $v_e$ is given by the formula $v_e = \sqrt{\frac{2GM}{R}}$. If the radius $R$ is doubled, the new escape speed will be $v_e’ = \sqrt{\frac{2GM}{2R}} = \sqrt{\frac{1}{2}}\sqrt{\frac{2GM}{R}} = \frac{v_e}{\sqrt{2}}$. Therefore, the escape speed decreases by a factor of $\sqrt{2}$.

Key Concept: Basic Definition

d) 11.2 km/s

[Solution Description] The escape speed from the surface of the Earth is a known value, approximately 11.2 km/s. This is the minimum speed required for an object to overcome Earth’s gravitational pull without additional propulsion.

Key Concept: Basic Concept

c) 11.2 km/s

[Solution Description] The escape speed from the Earth’s surface can be calculated using the formula for escape speed, but it is a known value. It is approximately 11.2 km/s.

Key Concept: Basic Calculation

d) 156.8 J

[Solution Description]

The work done in moving a particle between two heights is given by:

$W_{12} = mg(h_2 – h_1)$

Plugging in the values:

$$m &= 4 \, \text{kg},$$

$g = 9.8 \, \text{m/s}^2,$

$h_1 = 2 \, \text{m},$

$h_2 = 6 \, \text{m}.$

We get:

$$W_{12} &= 4 \times 9.8 \times (6 – 2)$$

$$= 4 \times 9.8 \times 4$$

$$= 156.8 \, \text{J}$$

Key Concept: Basic Definition

a) Zero

[Solution Description]

At an infinite distance from any mass, there are no forces acting between objects because they are infinitely separated. Hence, the gravitational potential energy is defined to be zero at infinity.

Key Concept: Basic Formula Application

c) 490 J

[Solution Description]

To find the gravitational potential energy near the Earth’s surface, we use the formula $W(h) = mgh + Wo$. Assuming $Wo = 0$, we have:

$W(h) = mgh = 10 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 5 \, \text{m} = 490 \, \text{J}$

Key Concept: Basic Formula

b) $Wo$

[Solution Description] The gravitational potential energy at the Earth’s surface is given by $W(h=0) = Wo$ according to the formula for gravitational potential energy near Earth’s surface.

Key Concept: Basic Formula Application

c) 88.2 Joules

[Solution Description]

The formula for work done in moving an object between two heights is given by:

$W_{12} = mg(h_2 – h_1)$

where $m = 3 \, \text{kg}$, $g = 9.8 \, \text{m/s}^2$, $h_1 = 1 \, \text{m}$, and $h_2 = 4 \, \text{m}$.

Substitute these values into the formula:

$W_{12} = 3 \times 9.8 \times (4 – 1) = 3 \times 9.8 \times 3 = 88.2 \, \text{Joules}$

Thus, the work done is $88.2 \, \text{Joules}$.

Key Concept: Basic Calculation

d) 490 Joules

[Solution Description]

The formula for gravitational potential energy near the Earth’s surface is $W(h) = mgh + W_0$. Here, $m = 5 \, \text{kg}$, $g = 9.8 \, \text{m/s}^2$, and $h = 10 \, \text{meters}$. We will assume $W_0 = 0$ for simplicity.
Calculate $W(h)$:
$W(h) = 5 \times 9.8 \times 10 = 490 \, \text{Joules}$
So, the gravitational potential energy is 490 Joules.

Key Concept: Basic Formula Application

b) $g(d) = g(1 – \frac{d}{R_E})$

[Solution Description] The formula for acceleration due to gravity at a depth $d$ below the Earth’s surface is given by $g(d) = g(1 – \frac{d}{R_E})$, where $g$ is the acceleration due to gravity on the Earth’s surface, and $R_E$ is the radius of the Earth.

Key Concept: Height Effect

b) Decreases

[Solution Description] As per the formula $g(h) = g(1 – \frac{2h}{R})$, the value of $g$ decreases as height $h$ increases since the fraction $\frac{2h}{R}$ becomes significant with an increase in height.

Key Concept: Gravitational Force Calculation

b) $3.71 \times 10^{-9}$ N

[Solution Description]

To find the gravitational force $F$, we use the equation:

$F = \frac{G M m}{d^2}$

where $G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2$, $M = 5 \, \text{kg}$, $m = 1 \, \text{kg}$, and $d = 0.3 \, \text{m}$.

Substitute the values into the formula:

$F = \frac{6.67 \times 10^{-11} \times 5 \times 1}{(0.3)^2}$

Calculate the denominator:

$(0.3)^2 = 0.09$

Now calculate the entire fraction:

$F = \frac{3.335 \times 10^{-10}}{0.09} = 3.705 \times 10^{-9} \, \text{N}$

Therefore, the force is approximately $3.71 \times 10^{-9} \, \text{N}$.

Key Concept: Historical Context

c) Henry Cavendish

[Solution Description] Henry Cavendish was the scientist who first determined the value of the gravitational constant $G$ experimentally in 1798.

Key Concept: Spherical Shells

b) As if the entire mass is concentrated at the center

[Solution Description] According to Newton’s shell theorem, if a point mass $m$ is located outside a hollow spherical shell, the gravitational force exerted by the shell on the mass is as if all of the shell’s mass $M$ were concentrated at its center. Therefore, the force is $F = \frac{G M m}{r^2}$, where $r$ is the distance from the center of the shell to the mass $m$.

Key Concept: Point Masses

b) Point masses

[Solution Description] Newton’s law of gravitation is ideally applied to point masses because it assumes that the entire mass of an object is concentrated at a single point. For extended bodies, this law can still be applied by considering the center of mass if the body is spherically symmetric or using calculus for more complex shapes.

Key Concept: Application of Proportionality

c) 72 years

[Solution Description] According to Kepler’s Third Law:
$T_1^2 / a_1^3 = T_2^2 / a_2^3$.
Given $a_1 = 15$, $T_1 = 9$, $a_2 = 60$. Solving for $T_2$:
$(9)^2 / (15)^3 = T_2^2 / (60)^3$
$81 / 3375 = T_2^2 / 216000$
$T_2^2 = (81 \cdot 216000) / 3375$
$T_2^2 = 5184$
$T_2 = \sqrt{5184} = 72$
The new time period is 72 years.

Key Concept: Perihelion and Aphelion

b) $$v_p > v_A$$

[Solution Description] According to the conservation of angular momentum for planetary motion, we have:

$$m r_p v_p = m r_A v_A$$

Since mass $$m$$ cancels out, the relationship between velocities and distances is given by:

$$r_p v_p = r_A v_A$$

Therefore, the correct answer is option b) $$v_p > v_A$$ because as the distance increases towards aphelion, the velocity decreases.

Key Concept: Application of Formula

b) 12 square units

[Solution Description]

According to Kepler’s second law, the line joining a planet and the Sun sweeps out equal areas in equal intervals of time. Given that the planet sweeps an area of 6 square units in 12 days, we can find the rate at which area is swept:

$$\text{Rate} = \frac{\Delta A}{\Delta t} = \frac{6 \text{ sq units}}{12 \text{ days}} = 0.5 \text{ sq units per day}$$.

Now, to find the area swept in 24 days, multiply the rate by the number of days:

$\Delta A_{24} = 0.5 \times 24 = 12 \text{ square units}.$

So, the planet will sweep out an area of 12 square units in 24 days.

Key Concept: Equal Areas Concept

d) 300 square units

[Solution Description]

According to Kepler’s Second Law, equal areas are swept in equal times. Therefore, if 150 square units are swept in 6 days, then:

$$\text{Area in 12 days} = 2 \times 150 = 300 \text{ square units}$$

Hence, the planet will sweep out 300 square units in 12 days.

Key Concept: Law of Areas

c) The areas swept are equal

[Solution Description] Kepler’s Second Law states that a line joining a planet and the Sun sweeps out equal areas in equal times. This implies that regardless of the length of time periods compared (whether they are long or short), the areas swept out during those periods are equal if the time intervals are equal. Therefore, for any given same-duration period, the areas will be the same irrespective of the specific time of year (winter or summer).

Key Concept: Semi-major Axis Calculation

c) 16 units

[Solution Description]

The formula for the relationship between the semi-major axis $a$, the foci distance $2c$, and the eccentricity $e$ is:

$e = \frac{c}{a}$

Given that $2c = 16$ units, we can find $c$ as:

$c = \frac{16}{2} = 8 \text{ units}$

Substituting $e = 0.5$ and $c = 8$ into the equation, we have:

$0.5 = \frac{8}{a}$

Solving for $a$, we get:

$a = \frac{8}{0.5} = 16 \text{ units}$

Thus, the length of the semi-major axis is 16 units.

Key Concept: Conceptual Understanding

c) $\frac{R_E}{2}$

[Solution Description] Given the condition $\frac{M_s}{M_E} = \left(\frac{R_E – d}{R_E}\right)^3 = \frac{1}{8}$, solving for $d$:

$\left(\frac{R_E – d}{R_E}\right)^3 = \frac{1}{8}$

Taking cube root of both sides,

$\frac{R_E – d}{R_E} = \frac{1}{2}$

Solving for $d$, we get

$R_E – d = \frac{R_E}{2} \implies d = \frac{R_E}{2}$

Key Concept: Proportionality Understanding

d) It is quartered

[Solution Description] Again using Kepler’s third law: $T^2 \propto a^3$. Initially, let $T_1 = 8$ and $T_2 = 1$. Then $\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{a_2}{a_1}\right)^3$. Hence, $\left(\frac{1}{8}\right)^2 = \left(\frac{a_2}{a_1}\right)^3$, resulting in $a_2/a_1 = \left(1/64\right)^{1/3} = 1/4$. The semi-major axis is quartered.

Key Concept: Ellipse Characteristics

a) Aphelion

[Solution Description]

In the context of an elliptical orbit, the point that is farthest from the Sun is known as the aphelion. The perihelion, conversely, is the closest point to the Sun.

Key Concept: Simple Calculation

c) 64

[Solution Description] By Kepler’s Third Law, $T^2 \propto a^3$, where $T$ is the period and $a$ is the semi-major axis. If $a = 4$, then:

$T^2 = 4^3 = 64$

Therefore, the square of the orbital period is 64 Earth years.

Key Concept: Theoretical Implications, Inverse Square Law

c) The gravitational force quadruples.

[Solution Description]

According to Newton’s law of universal gravitation, the gravitational force between two masses is inversely proportional to the square of the distance between them:

$F = \frac{G m_1 m_2}{r^2}$

When the distance $r$ is halved, $r’ = \frac{r}{2}$, the new gravitational force $F’$ becomes:

$F’ = \frac{G m_1 m_2}{(r/2)^2} = \frac{G m_1 m_2}{r^2/4} = 4 \times \frac{G m_1 m_2}{r^2} = 4F$

Thus, the gravitational force increases by a factor of four.

Key Concept: Integration with Newton’s Laws, Real-World Application

d) Less than an Earth year by a factor of $\sqrt{2}$

[Solution Description] According to Kepler’s third law, for planets orbiting the same star:

$T^2 \propto a^3/M$

where $M$ is the mass of the star. For Earth, orbiting our Sun:

$T_\text{earth}^2 \propto a^3/M_\text{sun}$

For the hypothetical star which has mass $2M_\text{sun}$:

$T_\text{hypo}^2 \propto a^3/(2M_\text{sun})$

Comparing both,

$\frac{T_\text{hypo}^2}{T_\text{earth}^2} = \frac{1}{2}$

So,

$T_\text{hypo} = \frac{T_\text{earth}}{\sqrt{2}}$

Hence, the orbital period is reduced by a factor of $\sqrt{2}$.

Key Concept: Theoretical Implications, Real World Application

c) 2.83 years

[Solution Description] Kepler’s Law of Periods states that the square of the orbital period ($T$) is proportional to the cube of the semi-major axis of its orbit ($a$):

$T^2 \propto a^3$

For Earth, $a_1 = 1$ AU and $T_1 = 1$ year. For the hypothetical planet, let $a_2 = 2$ AU. Then,

$\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{a_2}{a_1}\right)^3$

Substituting the known values,

$\left(\frac{T_2}{1}\right)^2 = (2)^3 = 8$

Solving for $T_2$,

$T_2 = \sqrt{8} \approx 2.83 \text{ years}$

This demonstrates the non-linear relationship between distance and period in the heliocentric model, fundamentally altering our comprehension of celestial mechanics.

Key Concept: Experimental Critique, Advanced Reasoning

b) Friction between the object and the plane

[Solution Description] Galileo used inclined planes to slow down the motion of falling objects so their movement could be measured more accurately. One potential source of error in these experiments is friction between the object and the surface of the plane. Friction would oppose the motion, effectively reducing the net acceleration observed compared to the true gravitational acceleration. Therefore, if friction isn’t adequately accounted for, it introduces errors into the calculated value of gravitational acceleration.

Key Concept: Historical Impact, Kepler’s Laws Derivation

c) They provided data supporting elliptical orbits

[Solution Description]

Tycho Brahe’s meticulous astronomical observations were crucial in helping Johannes Kepler derive his laws of planetary motion. Brahe’s data showed slight but significant discrepancies in the positions of planets compared to predictions based on circular orbits.

When Kepler analyzed this data, he found that these deviations could be explained if the orbits were ellipses rather than circles. This realization led to Kepler’s first law, the Law of Orbits, stating that planets move in elliptical orbits with the Sun at one focus.

Without Brahe’s accurate measurements, Kepler might not have been able to identify and mathematically prove the elliptical nature of planetary orbits.

Key Concept: Experimental Understanding, Multi-Concept Problems

c) Accurate angles determine gravitational force.

[Solution Description]

In Cavendish’s experiment, two lead spheres were used: large fixed ones and smaller suspended ones capable of twisting due to gravitational attraction. The angle through which the torsion balance twisted gave a measure of the tiny gravitational force between the masses.

The gravitational force $F$ can be expressed by:

$F = \frac{Gm_1m_2}{r^2}$

The torque $\tau$ on the wire is related to the force:

$\tau = k \theta$

Here, $\theta$ is the angular displacement measured, and $k$ is the torsional constant of the wire.

By substituting for $F$ in terms of torque:

$\frac{Gm_1m_2}{r^2} = k \theta$

Rearranging gives:

$G = \frac{k r^2 \theta}{m_1m_2}$

Hence, accurate measurement of $\theta$ directly impacts the precision of $G$, given known values of $k$, $r$, $m_1$, and $m_2$.

Key Concept: Theoretical Implications, Mathematical Application

b) It explains that gravitational force balances the satellite’s inertia, enabling circular orbits.

[Solution Description]

Newton’s Law of Universal Gravitation states that every mass attracts every other mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This gravitational pull provides the necessary centripetal force to keep satellites in a stable orbit. The satellite’s velocity must be just right – not too slow (or it will fall back to Earth) and not too fast (or it will escape Earth’s gravity).

Key Concept: Kepler’s Third Law

c) The square of the orbital period is proportional to the cube of the semi-major axis.

[Solution Description] According to Kepler’s third law, for any planet orbiting the Sun, the square of its orbital period $(T^2)$ is proportional to the cube of the semi-major axis of its orbit $(a^3)$. Mathematically, this can be expressed as $T^2 \propto a^3$. This means that if you double the distance from the Sun (semi-major axis), the period of rotation increases by a factor related to the cube root.