Grade 11 : Physics - Gravitation (Set 4)

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Grade 11 : Physics - Gravitation (Set 4)— Questions & Detailed Solutions

Two satellites, Satellite A and Satellite B, move around a planet in concentric circular orbits. The orbital radius of Satellite A is $R$, while the orbital radius of Satellite B is $4R$.
What is the ratio of the orbital time period of Satellite B to that of Satellite A ($T_B / T_A$)?

(A)

(B)

(C)

(D)

View Correct Answer & Detailed Solution

✓ Correct Option: (C)

Explanation:By Kepler's Third Law of Planetary Motion, $T^2 \propto r^3$, which implies $T \propto r^{3/2}$.
$$\frac{T_B}{T_A} = \left(\frac{R_B}{R_A}\right)^{3/2} = \left(\frac{4R}{R}\right)^{3/2} = 4^{3/2} = 8$$Thus, the ratio of the time periods is 8.

A research probe is elevated vertically from the Earth's surface to a height equal to the Earth's radius $R$.
If $g$ is the acceleration due to gravity at the surface of the Earth, what is the acceleration due to gravity at this height?

(A)

$g/2$

(B)

$g/4$

(C)

$g/8$

(D)

$g/16$

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation:The variation of acceleration due to gravity with height $h$ above the Earth's surface is given by:$$g' = g\left(\frac{R}{R+h}\right)^2$$For $h = R$:$$g' = g\left(\frac{R}{R+R}\right)^2 = g\left(\frac{1}{2}\right)^2 = \frac{g}{4}$$Hence, the acceleration due to gravity decreases to $g/4$.

A shaft is bored straight towards the center of the Earth down to a depth equal to half of the Earth's radius ($d = R/2$).
What is the acceleration due to gravity at this depth in terms of the surface gravity $g$?

(A)

$g/2$

(B)

$g/4$

(C)

$3g/4$

(D)

$g/3$

View Correct Answer & Detailed Solution

✓ Correct Option: (A)

Explanation:The acceleration due to gravity at a depth $d$ below the Earth's surface is:$$g' = g\left(1 - \frac{d}{R}\right)$$Substituting $d = \frac{R}{2}$:$$g' = g\left(1 - \frac{R/2}{R}\right) = g\left(1 - \frac{1}{2}\right) = \frac{g}{2}$$Thus, the gravity at depth $R/2$ is $g/2$.

Consider a hypothetical planet with twice the mass of Earth ($M' = 2M$) and half the radius of Earth ($R' = R/2$).
If $v_e$ is the escape velocity from the surface of Earth, what is the escape velocity $v_e'$ from the surface of this hypothetical planet?

(A)

$v_e$

(B)

$\sqrt{2}v_e$

(C)

$\sqrt{8}v_e$

(D)

$2v_e$

View Correct Answer & Detailed Solution

✓ Correct Option: (D)

Explanation:Escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.
For the new planet:$$v_e' = \sqrt{\frac{2G(2M)}{R/2}} = \sqrt{4 \cdot \frac{2GM}{R}} = 2\sqrt{\frac{2GM}{R}} = 2v_e$$The escape velocity from this planet is $2v_e$.

A weather satellite is placed in a circular orbit around Earth at an altitude of $h = 3R$ above the surface, where $R$ is the radius of the Earth.
What is the orbital speed of the satellite in terms of $G$, Earth's mass $M$, and Earth's radius $R$?

(A)

$\sqrt{\frac{GM}{R}}$

(B)

$\frac{1}{2}\sqrt{\frac{GM}{R}}$

(C)

$\frac{1}{3}\sqrt{\frac{GM}{R}}$

(D)

$\frac{1}{4}\sqrt{\frac{GM}{R}}$

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation:The orbital radius of the satellite is $r = R + h = R + 3R = 4R$.
The orbital speed formula is:$$v_o = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{4R}} = \frac{1}{2}\sqrt{\frac{GM}{R}}$$Therefore, the orbital speed is $\frac{1}{2}\sqrt{\frac{GM}{R}}$.

What is the total mechanical energy of a satellite of mass $m$ orbiting a planet of mass $M$ in a circular orbit of radius $r$?

(A)

$-\frac{GMm}{r}$

(B)

$\frac{GMm}{2r}$

(C)

$-\frac{GMm}{2r}$

(D)

$-\frac{2GMm}{r}$

View Correct Answer & Detailed Solution

✓ Correct Option: (C)

Explanation:Kinetic energy of the satellite is $K = \frac{GMm}{2r}$.
Potential energy of the satellite is $U = -\frac{GMm}{r}$.
Total mechanical energy $E = K + U$:$$E = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$$The total energy is $-\frac{GMm}{2r}$.

Two point masses $m_1 = M$ and $m_2 = 4M$ are fixed at a distance $D$ apart in space.
At what distance from mass $M$ along the line joining them is the net gravitational field intensity equal to zero?

(A)

$D/2$

(B)

$D/3$

(C)

$D/4$

(D)

$2D/3$

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✓ Correct Option: (B)

Explanation:Let the neutral point be at distance $x$ from mass $M$. The distance from mass $4M$ is $D - x$.
Equating gravitational field magnitudes:$$\frac{GM}{x^2} = \frac{G(4M)}{(D - x)^2} \implies \frac{1}{x^2} = \frac{4}{(D - x)^2}$$Taking square roots on both sides:$$\frac{1}{x} = \frac{2}{D - x} \implies D - x = 2x \implies 3x = D \implies x = \frac{D}{3}$$The zero-field point is at distance $D/3$ from $M$.

Planet Mass (in $M_E$) Radius (in $R_E$)
Planet X 2 1
Planet Y 4 2
Using the data provided in the table, determine the ratio of the gravitational acceleration on Planet X to that on Planet Y ($g_X / g_Y$).

Planet	Mass (in $M_E$)	Radius (in $R_E$)
Planet X	2	1
Planet Y	4	2

(A)

$2 : 1$

(B)

$1 : 1$

(C)

$4 : 1$

(D)

$1 : 2$

View Correct Answer & Detailed Solution

✓ Correct Option: (A)

Explanation:Surface gravity formula is $g = \frac{GM}{R^2}$.
For Planet X: $g_X = \frac{G(2M_E)}{(R_E)^2} = 2 g_E$
For Planet Y: $g_Y = \frac{G(4M_E)}{(2R_E)^2} = \frac{4}{4} g_E = 1 g_E$
Ratio:$$\frac{g_X}{g_Y} = \frac{2 g_E}{1 g_E} = \frac{2}{1}$$Thus, the ratio $g_X : g_Y = 2 : 1$.

A straight tunnel is drilled completely through the center of a uniform solid Earth of mass $M$ and radius $R$. A particle is released from rest at one end of the tunnel.
Assuming no friction, what is the period of Simple Harmonic Motion (SHM) performed by the particle inside the tunnel?

(A)

$\pi \sqrt{\frac{R}{g}}$

(B)

$2\pi \sqrt{\frac{R}{g}}$

(C)

$2\pi \sqrt{\frac{2R}{g}}$

(D)

$4\pi \sqrt{\frac{R}{g}}$

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation:Inside a uniform solid sphere, restoring force at distance $x$ from center is $F = -\frac{GMm}{R^3}x = -m\left(\frac{g}{R}\right)x$.
The angular frequency of oscillation is $\omega = \sqrt{\frac{g}{R}}$.
The time period is:$$T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{R}{g}}$$Thus, the period of oscillation is $2\pi \sqrt{R/g}$.

Q10

A comet moves around the Sun in an elliptical orbit. Its distance at perihelion (closest approach) is $r_1 = r$ and at aphelion (farthest distance) is $r_2 = 3r$.
What is the ratio of the linear speed of the comet at perihelion ($v_1$) to its linear speed at aphelion ($v_2$)?

(A)

$1 : 3$

(B)

$1 : 9$

(C)

$3 : 1$

(D)

$9 : 1$

View Correct Answer & Detailed Solution

✓ Correct Option: (C)

Explanation:By conservation of angular momentum about the Sun:$$m v_1 r_1 = m v_2 r_2 \implies v_1 r_1 = v_2 r_2$$Given $r_1 = r$ and $r_2 = 3r$:$$\frac{v_1}{v_2} = \frac{r_2}{r_1} = \frac{3r}{r} = 3$$Thus, $v_1 : v_2 = 3 : 1$.

Q11

How much work must be done by an external agent to raise a mass $m$ slowly from the surface of Earth (radius $R$) to a height $h = 2R$?

(A)

$2 mgR$

(B)

$\frac{1}{2} mgR$

(C)

$\frac{2}{3} mgR$

(D)

$\frac{3}{4} mgR$

View Correct Answer & Detailed Solution

✓ Correct Option: (C)

Explanation:The work done equals change in potential energy:$$W = \Delta U = U_f - U_i = -\frac{GMm}{R+h} - \left(-\frac{GMm}{R}\right) = \frac{GMm h}{R(R+h)}$$Substituting $GM = gR^2$ and $h = 2R$:$$W = \frac{gR^2 m (2R)}{R(3R)} = \frac{2}{3} mgR$$The required work done is $\frac{2}{3} mgR$.

Q12

An astronaut inside an orbiting space station drops an apple. The space station is in a circular orbit around Earth.
What is the apparent weight of the apple as measured by the astronaut inside the station?

(A)

Zero

(B)

Equal to mg

(C)

Greater than mg

(D)

Equal to half of mg

View Correct Answer & Detailed Solution

✓ Correct Option: (A)

Explanation:Both the astronaut and the apple inside the satellite experience identical gravitational acceleration toward Earth's center. As a result, there is no normal reaction force between the object and the satellite environment, rendering the apparent weight zero (weightlessness).

Q13

Taking Earth's rotation into account with angular velocity $\omega$ and radius $R$, what is the effective acceleration due to gravity $g'$ at the equator?

(A)

$g$

(B)

$g + \omega^2 R$

(C)

$g - \frac{1}{2}\omega^2 R$

(D)

$g - \omega^2 R$

View Correct Answer & Detailed Solution

✓ Correct Option: (D)

Explanation:The formula for acceleration due to gravity at latitude $\lambda$ considering rotation is:$$g' = g - \omega^2 R \cos^2\lambda$$At the equator, latitude $\lambda = 0^\circ$, so $\cos(0^\circ) = 1$.
Thus, $g' = g - \omega^2 R$.

Q14

What is the ratio of the escape velocity from Earth's surface to the orbital velocity of a satellite orbiting very close to Earth's surface?

(A)

$1 : 1$

(B)

$\sqrt{2} : 1$

(C)

$2 : 1$

(D)

$1 : \sqrt{2}$

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation:Escape velocity near Earth's surface: $v_e = \sqrt{\frac{2GM}{R}}$.
Orbital velocity near Earth's surface: $v_o = \sqrt{\frac{GM}{R}}$.
The ratio is:$$\frac{v_e}{v_o} = \frac{\sqrt{\frac{2GM}{R}}}{\sqrt{\frac{GM}{R}}} = \sqrt{2}$$Hence, $v_e : v_o = \sqrt{2} : 1$.

Q15

A satellite orbiting Earth in a circular orbit of radius $r$ encounters thin atmospheric resistance, causing its orbit radius to decrease to $r/2$.
How does the kinetic energy of the satellite change after moving into this lower circular orbit?

(A)

Decreases by half

(B)

Remains unchanged

(C)

Doubles

(D)

Quadruples

View Correct Answer & Detailed Solution

✓ Correct Option: (C)

Explanation:Kinetic energy in a circular orbit of radius $r$ is $K_1 = \frac{GMm}{2r}$.
For new radius $r' = r/2$:$$K_2 = \frac{GMm}{2(r/2)} = \frac{GMm}{r} = 2 K_1$$The kinetic energy doubles as the satellite moves closer to the center.

Q16

What is the gravitational potential at the center of a solid uniform sphere of mass $M$ and radius $R$ (taking potential at infinity as zero)?

(A)

$-\frac{3GM}{2R}$

(B)

$-\frac{GM}{R}$

(C)

$-\frac{GM}{2R}$

(D)

Zero

View Correct Answer & Detailed Solution

✓ Correct Option: (A)

Explanation:The potential at a distance $r \le R$ inside a uniform solid sphere is given by:$$V(r) = -\frac{GM}{2R^3}(3R^2 - r^2)$$At the center ($r = 0$):$$V_c = -\frac{GM}{2R^3}(3R^2) = -\frac{3GM}{2R}$$The potential at the center is $-\frac{3GM}{2R}$ (1.5 times the potential at the surface).

Q17

For small distances above and below Earth's surface ($h, d \ll R$), the acceleration due to gravity decreases at both height $h$ and depth $d$.
At what height $h$ above Earth's surface is the decrease in $g$ equal to the decrease in $g$ at a depth of $d = 20km$ below the surface?

(A)

20 km

(B)

10 km

(C)

40 km

(D)

5 km

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation:For small $h \ll R$, fractional decrease in $g$ at height $h$ is $\frac{\Delta g}{g} = \frac{2h}{R}$.
At depth $d$, fractional decrease in $g$ is $\frac{\Delta g}{g} = \frac{d}{R}$.
Equating both decreases:$$\frac{2h}{R} = \frac{d}{R} \implies 2h = d \implies h = \frac{d}{2}$$For $d = 20km$, $h = 10km$.

Q18

Two identical stars, each of mass $M$, move in circular orbits around their common center of mass under their mutual gravitational attraction. The distance between the centers of the two stars is $2R$.
What is the time period $T$ of revolution of each star?

(A)

$2\pi \sqrt{\frac{R^3}{GM}}$

(B)

$2\pi \sqrt{\frac{2R^3}{GM}}$

(C)

$4\pi \sqrt{\frac{R^3}{GM}}$

(D)

$4\pi \sqrt{\frac{2R^3}{GM}}$

View Correct Answer & Detailed Solution

✓ Correct Option: (C)

Explanation:Each star moves in a circle of radius $R$ about center of mass.
Gravitational force $F = \frac{G M^2}{(2R)^2} = \frac{GM^2}{4R^2}$.
Centripetal force equation:$$\frac{GM^2}{4R^2} = M \omega^2 R \implies \omega^2 = \frac{GM}{4R^3} \implies \omega = \frac{1}{2}\sqrt{\frac{GM}{R^3}}$$Time period $T = \frac{2\pi}{\omega} = 4\pi \sqrt{\frac{R^3}{GM}}$.

Q19

A body is projected vertically upward from Earth's surface with an initial velocity equal to half of the escape velocity ($v = \frac{1}{2}v_e$).
What is the maximum height $h$ attained by the body above Earth's surface in terms of Earth's radius $R$?

(A)

$R/2$

(B)

$R/3$

(C)

$R/4$

(D)

$2R/3$

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation:Escape velocity $v_e = \sqrt{\frac{2GM}{R}}$, so $v = \frac{1}{2}\sqrt{\frac{2GM}{R}} = \sqrt{\frac{GM}{2R}}$.
By conservation of energy:$$\frac{1}{2}mv^2 - \frac{GMm}{R} = -\frac{GMm}{R+h}$$Substitute $v^2 = \frac{GM}{2R}$:$$\frac{1}{2}m\left(\frac{GM}{2R}\right) - \frac{GMm}{R} = -\frac{GMm}{R+h} \implies -\frac{3}{4}\frac{GMm}{R} = -\frac{GMm}{R+h}$$Hence $R+h = \frac{4}{3}R \implies h = \frac{R}{3}$.

Q20

Statement I: The gravitational force between two point masses is independent of the presence of other bodies and the intervening medium.
Statement II: Gravitational force can be shielded by surrounding a mass with a thick shell of lead.
Which of the following statements is correct regarding the assertions above?

(A)

Both Statement I and Statement II are true.

(B)

Both Statement I and Statement II are false.

(C)

Statement I is true, but Statement II is false.

(D)

Statement I is false, but Statement II is true.

View Correct Answer & Detailed Solution

✓ Correct Option: (C)

Explanation:Statement I is true because Newton's Law of Gravitation is independent of any surrounding medium or additional masses.
Statement II is false because gravitational forces cannot be shielded or absorbed by any material medium or barrier.

grade 11physicsgravitationkeplers lawsescape velocityorbital velocitygravitational potential

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Grade 11 : Physics - Gravitation (Set 4)

Grade 11 : Physics - Gravitation (Set 4)— Questions & Detailed Solutions

Two satellites, Satellite A and Satellite B, move around a planet in concentric circular orbits. The orbital radius of Satellite A is $R$, while the orbital radius of Satellite B is $4R$.What is the ratio of the orbital time period of Satellite B to that of Satellite A ($T_B / T_A$)?

A research probe is elevated vertically from the Earth's surface to a height equal to the Earth's radius $R$.If $g$ is the acceleration due to gravity at the surface of the Earth, what is the acceleration due to gravity at this height?

A shaft is bored straight towards the center of the Earth down to a depth equal to half of the Earth's radius ($d = R/2$).What is the acceleration due to gravity at this depth in terms of the surface gravity $g$?

Consider a hypothetical planet with twice the mass of Earth ($M' = 2M$) and half the radius of Earth ($R' = R/2$).If $v_e$ is the escape velocity from the surface of Earth, what is the escape velocity $v_e'$ from the surface of this hypothetical planet?

A weather satellite is placed in a circular orbit around Earth at an altitude of $h = 3R$ above the surface, where $R$ is the radius of the Earth.What is the orbital speed of the satellite in terms of $G$, Earth's mass $M$, and Earth's radius $R$?

What is the total mechanical energy of a satellite of mass $m$ orbiting a planet of mass $M$ in a circular orbit of radius $r$?

Two point masses $m_1 = M$ and $m_2 = 4M$ are fixed at a distance $D$ apart in space.At what distance from mass $M$ along the line joining them is the net gravitational field intensity equal to zero?

PlanetMass (in $M_E$)Radius (in $R_E$)Planet X21Planet Y42Using the data provided in the table, determine the ratio of the gravitational acceleration on Planet X to that on Planet Y ($g_X / g_Y$).

A straight tunnel is drilled completely through the center of a uniform solid Earth of mass $M$ and radius $R$. A particle is released from rest at one end of the tunnel.Assuming no friction, what is the period of Simple Harmonic Motion (SHM) performed by the particle inside the tunnel?

A comet moves around the Sun in an elliptical orbit. Its distance at perihelion (closest approach) is $r_1 = r$ and at aphelion (farthest distance) is $r_2 = 3r$.What is the ratio of the linear speed of the comet at perihelion ($v_1$) to its linear speed at aphelion ($v_2$)?

How much work must be done by an external agent to raise a mass $m$ slowly from the surface of Earth (radius $R$) to a height $h = 2R$?

An astronaut inside an orbiting space station drops an apple. The space station is in a circular orbit around Earth.What is the apparent weight of the apple as measured by the astronaut inside the station?

Taking Earth's rotation into account with angular velocity $\omega$ and radius $R$, what is the effective acceleration due to gravity $g'$ at the equator?

What is the ratio of the escape velocity from Earth's surface to the orbital velocity of a satellite orbiting very close to Earth's surface?

A satellite orbiting Earth in a circular orbit of radius $r$ encounters thin atmospheric resistance, causing its orbit radius to decrease to $r/2$.How does the kinetic energy of the satellite change after moving into this lower circular orbit?

What is the gravitational potential at the center of a solid uniform sphere of mass $M$ and radius $R$ (taking potential at infinity as zero)?

For small distances above and below Earth's surface ($h, d \ll R$), the acceleration due to gravity decreases at both height $h$ and depth $d$.At what height $h$ above Earth's surface is the decrease in $g$ equal to the decrease in $g$ at a depth of $d = 20km$ below the surface?

Two identical stars, each of mass $M$, move in circular orbits around their common center of mass under their mutual gravitational attraction. The distance between the centers of the two stars is $2R$.What is the time period $T$ of revolution of each star?

A body is projected vertically upward from Earth's surface with an initial velocity equal to half of the escape velocity ($v = \frac{1}{2}v_e$).What is the maximum height $h$ attained by the body above Earth's surface in terms of Earth's radius $R$?

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Two satellites, Satellite A and Satellite B, move around a planet in concentric circular orbits. The orbital radius of Satellite A is $R$, while the orbital radius of Satellite B is $4R$.
What is the ratio of the orbital time period of Satellite B to that of Satellite A ($T_B / T_A$)?

A research probe is elevated vertically from the Earth's surface to a height equal to the Earth's radius $R$.
If $g$ is the acceleration due to gravity at the surface of the Earth, what is the acceleration due to gravity at this height?

A shaft is bored straight towards the center of the Earth down to a depth equal to half of the Earth's radius ($d = R/2$).
What is the acceleration due to gravity at this depth in terms of the surface gravity $g$?

Consider a hypothetical planet with twice the mass of Earth ($M' = 2M$) and half the radius of Earth ($R' = R/2$).
If $v_e$ is the escape velocity from the surface of Earth, what is the escape velocity $v_e'$ from the surface of this hypothetical planet?

A weather satellite is placed in a circular orbit around Earth at an altitude of $h = 3R$ above the surface, where $R$ is the radius of the Earth.
What is the orbital speed of the satellite in terms of $G$, Earth's mass $M$, and Earth's radius $R$?

Two point masses $m_1 = M$ and $m_2 = 4M$ are fixed at a distance $D$ apart in space.
At what distance from mass $M$ along the line joining them is the net gravitational field intensity equal to zero?

Planet Mass (in $M_E$) Radius (in $R_E$)
Planet X 2 1
Planet Y 4 2
Using the data provided in the table, determine the ratio of the gravitational acceleration on Planet X to that on Planet Y ($g_X / g_Y$).

A straight tunnel is drilled completely through the center of a uniform solid Earth of mass $M$ and radius $R$. A particle is released from rest at one end of the tunnel.
Assuming no friction, what is the period of Simple Harmonic Motion (SHM) performed by the particle inside the tunnel?

A comet moves around the Sun in an elliptical orbit. Its distance at perihelion (closest approach) is $r_1 = r$ and at aphelion (farthest distance) is $r_2 = 3r$.
What is the ratio of the linear speed of the comet at perihelion ($v_1$) to its linear speed at aphelion ($v_2$)?

An astronaut inside an orbiting space station drops an apple. The space station is in a circular orbit around Earth.
What is the apparent weight of the apple as measured by the astronaut inside the station?

A satellite orbiting Earth in a circular orbit of radius $r$ encounters thin atmospheric resistance, causing its orbit radius to decrease to $r/2$.
How does the kinetic energy of the satellite change after moving into this lower circular orbit?

For small distances above and below Earth's surface ($h, d \ll R$), the acceleration due to gravity decreases at both height $h$ and depth $d$.
At what height $h$ above Earth's surface is the decrease in $g$ equal to the decrease in $g$ at a depth of $d = 20km$ below the surface?

Two identical stars, each of mass $M$, move in circular orbits around their common center of mass under their mutual gravitational attraction. The distance between the centers of the two stars is $2R$.
What is the time period $T$ of revolution of each star?

A body is projected vertically upward from Earth's surface with an initial velocity equal to half of the escape velocity ($v = \frac{1}{2}v_e$).
What is the maximum height $h$ attained by the body above Earth's surface in terms of Earth's radius $R$?