QBManager

Gravitation

138606 The distance of Neptune and Saturn from the sun is nearly $10^{13}$ and $10^{12}$ meter respectively. Assuming that they move in circular orbits, their periodic times will be in the ratio of

1 10

2 100

3

10 \sqrt{10}

4 1000

Explanation:

C Given that,

Now we using Kepler's third law,
$T^{2} \propto a^{3}$
$\frac{T_{N}^{2}}{T_{S}^{2}} = \frac{a_{N}^{3}}{a_{S}^{3}}$
${(\frac{T_{N}}{T_{S}})}^{2} = {(\frac{10^{13}}{10^{12}})}^{3} = 10^{3}$
$\frac{T_{N}}{T_{S}} = 10 \sqrt{10}$

AIPMT 1994

Gravitation

138609 Kepler's third law states that the square of period of revolution ( $T$ ) of a planet around the sun, is proportional to third power of average distance, $r$ between the sun and the planet i.e. $T^{2} = {Kr}^{3}$
Here, $K$ is constant.
If masses of the sun and the planet are $M$ and $m$ respectively, then as per Newton's law and $m$ respectively, force of attraction between them is $F = \frac{G M m}{r^{2}}$ , where $G$ is gravitational constant.
The relation between $G$ and $K$ is described as

1

GK = 4 π^{2}

2

GMK = 4 π^{2}

3

K = G

4

K = \frac{1}{G}

Explanation:

B Gravitational force of attraction between sun and planet provides centripetal force for the orbit of planet.
$∴ \frac{GMm}{r^{2}} = \frac{{mv}^{2}}{r} =$ Centripetal force
$v^{2} = \frac{GM}{r} \dots \dots$
Time period of the planet is given by
$T = \frac{2 π r}{v}$
Now, squaring both side
$T^{2} = \frac{4 π^{2} r^{2}}{v^{2}}$
$T^{2} = \frac{4 π^{2} r^{2}}{(\frac{GM}{r})}$
$T^{2} = \frac{4 π^{2} r^{3}}{GM} \dots \dots$
According to question
$T^{2} = {Kr}^{3}$
Comparing equation (ii) and (iii), we get
${Kr}^{3} = \frac{4 π^{2} r^{3}}{GM}$
$K = \frac{4 π^{2}}{GM}$ $$
$∴ GMK = 4 π^{2}$

BCECE-2017

Gravitation

138610 A planet moving around the sun sweeps area $A_{1}$ in 2 days, $A_{2}$ in 4 days and $A_{3}$ in 9 days. Then, relation between them.

1

A_{1} = A_{2} = A_{3}

2

9 A_{1} = 3 A_{2} = 2 A_{3}

3

18 A_{1} = 9 A_{2} = 4 A_{3}

4

3 A_{1} = 4 A_{2} = 6 A_{3}

Explanation:

C Given that,
$A_{1} = 2 days$
$A_{2} = 4 days$
$A_{3} = 9 days$
According to the Kepler's second law,
$\frac{dA}{dt} = \frac{L}{2 m} = constant$
Where, $L =$ Angular momentum of the planet
$m =$ Mass of the planet
$\frac{dA}{dt} = constant$
$\frac{A_{1}}{2} = constant$
$\frac{A_{2}}{4} = constant$
$\frac{A_{3}}{9} = constant$
According to question-
$\frac{A_{1}}{2} = \frac{A_{2}}{4} = \frac{A_{3}}{9}$
Now we taking LCM 2, 4, 9 is equal to 36
Therefore,
$36 \times \frac{A_{1}}{2} = 36 \times \frac{A_{2}}{4} = 36 \times \frac{A_{3}}{9}$
$18 A_{1} = 9 A_{2} = 4 A_{3}$

BCECE-2018

Gravitation

138611 Two planets revolves around the sun with frequencies $N_{1}$ and $N_{2}$ revolutions per year. If their average radii (orbital) be $R_{1}$ and $R_{2}$ respectively, then $R_{1} / R_{2}$ is equal to-

1

{(N_{1} / N_{2})}^{2 / 3}

2

{(N_{1} / N_{2})}^{3 / 2}

3

{(N_{2} / N_{1})}^{2 / 3}

4

{(N_{2} / N_{1})}^{3 / 2}

Explanation:

C
According to Kepler's law of planetary motion $T^{2} \propto R^{3}$
According to question -
$T_{1} = \frac{1}{N_{1}}, T_{2} = \frac{1}{N_{2}}$
$∴ \frac{T^{2}}{R^{3}} =$ constant
$\frac{T_{1}^{2}}{R_{1}^{3}} = \frac{T_{2}^{2}}{R_{2}^{3}}$
$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{R_{1}^{3}}{R_{2}^{3}}$
$\frac{R_{1}}{R_{2}} = {(\frac{T_{1}}{T_{2}})}^{2 / 3}$
$∴ \frac{R_{1}}{R_{2}} = {(\frac{N_{2}}{N_{1}})}^{2 / 3}$

BCECE-2012

Gravitation

138614 The earth takes $24 hr$ to rotate once about its axis. How much time does the sun take to shift by $5^{\circ}$ when viewed from the earth?

1

20 \min

2

15 \min

3

10 \min

4

5 \min

Explanation:

A The Earth takes 24 hours to complete one rotation around own axis and complete one revolution about sun in 365 days.
A complete rotation of earth makes $360^{\circ} = 24 h$
$1^{\circ} = \frac{24 h}{360}$
$1^{\circ} = \frac{24 \times 60 \min}{360}$
$1^{\circ} = 4 \min$
$∴$ For $5^{\circ} = 5 \times 4 \min$
$= 20 \min$

J and K-CET-2014

Gravitation

138606 The distance of Neptune and Saturn from the sun is nearly $10^{13}$ and $10^{12}$ meter respectively. Assuming that they move in circular orbits, their periodic times will be in the ratio of

1 10

2 100

3

10 \sqrt{10}

4 1000

Explanation:

C Given that,

Now we using Kepler's third law,
$T^{2} \propto a^{3}$
$\frac{T_{N}^{2}}{T_{S}^{2}} = \frac{a_{N}^{3}}{a_{S}^{3}}$
${(\frac{T_{N}}{T_{S}})}^{2} = {(\frac{10^{13}}{10^{12}})}^{3} = 10^{3}$
$\frac{T_{N}}{T_{S}} = 10 \sqrt{10}$

AIPMT 1994

Gravitation

138609 Kepler's third law states that the square of period of revolution ( $T$ ) of a planet around the sun, is proportional to third power of average distance, $r$ between the sun and the planet i.e. $T^{2} = {Kr}^{3}$
Here, $K$ is constant.
If masses of the sun and the planet are $M$ and $m$ respectively, then as per Newton's law and $m$ respectively, force of attraction between them is $F = \frac{G M m}{r^{2}}$ , where $G$ is gravitational constant.
The relation between $G$ and $K$ is described as

1

GK = 4 π^{2}

2

GMK = 4 π^{2}

3

K = G

4

K = \frac{1}{G}

Explanation:

B Gravitational force of attraction between sun and planet provides centripetal force for the orbit of planet.
$∴ \frac{GMm}{r^{2}} = \frac{{mv}^{2}}{r} =$ Centripetal force
$v^{2} = \frac{GM}{r} \dots \dots$
Time period of the planet is given by
$T = \frac{2 π r}{v}$
Now, squaring both side
$T^{2} = \frac{4 π^{2} r^{2}}{v^{2}}$
$T^{2} = \frac{4 π^{2} r^{2}}{(\frac{GM}{r})}$
$T^{2} = \frac{4 π^{2} r^{3}}{GM} \dots \dots$
According to question
$T^{2} = {Kr}^{3}$
Comparing equation (ii) and (iii), we get
${Kr}^{3} = \frac{4 π^{2} r^{3}}{GM}$
$K = \frac{4 π^{2}}{GM}$ $$
$∴ GMK = 4 π^{2}$

BCECE-2017

Gravitation

138610 A planet moving around the sun sweeps area $A_{1}$ in 2 days, $A_{2}$ in 4 days and $A_{3}$ in 9 days. Then, relation between them.

1

A_{1} = A_{2} = A_{3}

2

9 A_{1} = 3 A_{2} = 2 A_{3}

3

18 A_{1} = 9 A_{2} = 4 A_{3}

4

3 A_{1} = 4 A_{2} = 6 A_{3}

Explanation:

C Given that,
$A_{1} = 2 days$
$A_{2} = 4 days$
$A_{3} = 9 days$
According to the Kepler's second law,
$\frac{dA}{dt} = \frac{L}{2 m} = constant$
Where, $L =$ Angular momentum of the planet
$m =$ Mass of the planet
$\frac{dA}{dt} = constant$
$\frac{A_{1}}{2} = constant$
$\frac{A_{2}}{4} = constant$
$\frac{A_{3}}{9} = constant$
According to question-
$\frac{A_{1}}{2} = \frac{A_{2}}{4} = \frac{A_{3}}{9}$
Now we taking LCM 2, 4, 9 is equal to 36
Therefore,
$36 \times \frac{A_{1}}{2} = 36 \times \frac{A_{2}}{4} = 36 \times \frac{A_{3}}{9}$
$18 A_{1} = 9 A_{2} = 4 A_{3}$

BCECE-2018

Gravitation

138611 Two planets revolves around the sun with frequencies $N_{1}$ and $N_{2}$ revolutions per year. If their average radii (orbital) be $R_{1}$ and $R_{2}$ respectively, then $R_{1} / R_{2}$ is equal to-

1

{(N_{1} / N_{2})}^{2 / 3}

2

{(N_{1} / N_{2})}^{3 / 2}

3

{(N_{2} / N_{1})}^{2 / 3}

4

{(N_{2} / N_{1})}^{3 / 2}

Explanation:

C
According to Kepler's law of planetary motion $T^{2} \propto R^{3}$
According to question -
$T_{1} = \frac{1}{N_{1}}, T_{2} = \frac{1}{N_{2}}$
$∴ \frac{T^{2}}{R^{3}} =$ constant
$\frac{T_{1}^{2}}{R_{1}^{3}} = \frac{T_{2}^{2}}{R_{2}^{3}}$
$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{R_{1}^{3}}{R_{2}^{3}}$
$\frac{R_{1}}{R_{2}} = {(\frac{T_{1}}{T_{2}})}^{2 / 3}$
$∴ \frac{R_{1}}{R_{2}} = {(\frac{N_{2}}{N_{1}})}^{2 / 3}$

BCECE-2012

Gravitation

138614 The earth takes $24 hr$ to rotate once about its axis. How much time does the sun take to shift by $5^{\circ}$ when viewed from the earth?

1

20 \min

2

15 \min

3

10 \min

4

5 \min

Explanation:

A The Earth takes 24 hours to complete one rotation around own axis and complete one revolution about sun in 365 days.
A complete rotation of earth makes $360^{\circ} = 24 h$
$1^{\circ} = \frac{24 h}{360}$
$1^{\circ} = \frac{24 \times 60 \min}{360}$
$1^{\circ} = 4 \min$
$∴$ For $5^{\circ} = 5 \times 4 \min$
$= 20 \min$

J and K-CET-2014

Gravitation

138606 The distance of Neptune and Saturn from the sun is nearly $10^{13}$ and $10^{12}$ meter respectively. Assuming that they move in circular orbits, their periodic times will be in the ratio of

1 10

2 100

3

10 \sqrt{10}

4 1000

Explanation:

C Given that,

Now we using Kepler's third law,
$T^{2} \propto a^{3}$
$\frac{T_{N}^{2}}{T_{S}^{2}} = \frac{a_{N}^{3}}{a_{S}^{3}}$
${(\frac{T_{N}}{T_{S}})}^{2} = {(\frac{10^{13}}{10^{12}})}^{3} = 10^{3}$
$\frac{T_{N}}{T_{S}} = 10 \sqrt{10}$

AIPMT 1994

Gravitation

138609 Kepler's third law states that the square of period of revolution ( $T$ ) of a planet around the sun, is proportional to third power of average distance, $r$ between the sun and the planet i.e. $T^{2} = {Kr}^{3}$
Here, $K$ is constant.
If masses of the sun and the planet are $M$ and $m$ respectively, then as per Newton's law and $m$ respectively, force of attraction between them is $F = \frac{G M m}{r^{2}}$ , where $G$ is gravitational constant.
The relation between $G$ and $K$ is described as

1

GK = 4 π^{2}

2

GMK = 4 π^{2}

3

K = G

4

K = \frac{1}{G}

Explanation:

B Gravitational force of attraction between sun and planet provides centripetal force for the orbit of planet.
$∴ \frac{GMm}{r^{2}} = \frac{{mv}^{2}}{r} =$ Centripetal force
$v^{2} = \frac{GM}{r} \dots \dots$
Time period of the planet is given by
$T = \frac{2 π r}{v}$
Now, squaring both side
$T^{2} = \frac{4 π^{2} r^{2}}{v^{2}}$
$T^{2} = \frac{4 π^{2} r^{2}}{(\frac{GM}{r})}$
$T^{2} = \frac{4 π^{2} r^{3}}{GM} \dots \dots$
According to question
$T^{2} = {Kr}^{3}$
Comparing equation (ii) and (iii), we get
${Kr}^{3} = \frac{4 π^{2} r^{3}}{GM}$
$K = \frac{4 π^{2}}{GM}$ $$
$∴ GMK = 4 π^{2}$

BCECE-2017

Gravitation

138610 A planet moving around the sun sweeps area $A_{1}$ in 2 days, $A_{2}$ in 4 days and $A_{3}$ in 9 days. Then, relation between them.

1

A_{1} = A_{2} = A_{3}

2

9 A_{1} = 3 A_{2} = 2 A_{3}

3

18 A_{1} = 9 A_{2} = 4 A_{3}

4

3 A_{1} = 4 A_{2} = 6 A_{3}

Explanation:

C Given that,
$A_{1} = 2 days$
$A_{2} = 4 days$
$A_{3} = 9 days$
According to the Kepler's second law,
$\frac{dA}{dt} = \frac{L}{2 m} = constant$
Where, $L =$ Angular momentum of the planet
$m =$ Mass of the planet
$\frac{dA}{dt} = constant$
$\frac{A_{1}}{2} = constant$
$\frac{A_{2}}{4} = constant$
$\frac{A_{3}}{9} = constant$
According to question-
$\frac{A_{1}}{2} = \frac{A_{2}}{4} = \frac{A_{3}}{9}$
Now we taking LCM 2, 4, 9 is equal to 36
Therefore,
$36 \times \frac{A_{1}}{2} = 36 \times \frac{A_{2}}{4} = 36 \times \frac{A_{3}}{9}$
$18 A_{1} = 9 A_{2} = 4 A_{3}$

BCECE-2018

Gravitation

138611 Two planets revolves around the sun with frequencies $N_{1}$ and $N_{2}$ revolutions per year. If their average radii (orbital) be $R_{1}$ and $R_{2}$ respectively, then $R_{1} / R_{2}$ is equal to-

1

{(N_{1} / N_{2})}^{2 / 3}

2

{(N_{1} / N_{2})}^{3 / 2}

3

{(N_{2} / N_{1})}^{2 / 3}

4

{(N_{2} / N_{1})}^{3 / 2}

Explanation:

C
According to Kepler's law of planetary motion $T^{2} \propto R^{3}$
According to question -
$T_{1} = \frac{1}{N_{1}}, T_{2} = \frac{1}{N_{2}}$
$∴ \frac{T^{2}}{R^{3}} =$ constant
$\frac{T_{1}^{2}}{R_{1}^{3}} = \frac{T_{2}^{2}}{R_{2}^{3}}$
$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{R_{1}^{3}}{R_{2}^{3}}$
$\frac{R_{1}}{R_{2}} = {(\frac{T_{1}}{T_{2}})}^{2 / 3}$
$∴ \frac{R_{1}}{R_{2}} = {(\frac{N_{2}}{N_{1}})}^{2 / 3}$

BCECE-2012

Gravitation

138614 The earth takes $24 hr$ to rotate once about its axis. How much time does the sun take to shift by $5^{\circ}$ when viewed from the earth?

1

20 \min

2

15 \min

3

10 \min

4

5 \min

Explanation:

A The Earth takes 24 hours to complete one rotation around own axis and complete one revolution about sun in 365 days.
A complete rotation of earth makes $360^{\circ} = 24 h$
$1^{\circ} = \frac{24 h}{360}$
$1^{\circ} = \frac{24 \times 60 \min}{360}$
$1^{\circ} = 4 \min$
$∴$ For $5^{\circ} = 5 \times 4 \min$
$= 20 \min$

J and K-CET-2014

Gravitation

138606 The distance of Neptune and Saturn from the sun is nearly $10^{13}$ and $10^{12}$ meter respectively. Assuming that they move in circular orbits, their periodic times will be in the ratio of

1 10

2 100

3

10 \sqrt{10}

4 1000

Explanation:

C Given that,

Now we using Kepler's third law,
$T^{2} \propto a^{3}$
$\frac{T_{N}^{2}}{T_{S}^{2}} = \frac{a_{N}^{3}}{a_{S}^{3}}$
${(\frac{T_{N}}{T_{S}})}^{2} = {(\frac{10^{13}}{10^{12}})}^{3} = 10^{3}$
$\frac{T_{N}}{T_{S}} = 10 \sqrt{10}$

AIPMT 1994

Gravitation

138609 Kepler's third law states that the square of period of revolution ( $T$ ) of a planet around the sun, is proportional to third power of average distance, $r$ between the sun and the planet i.e. $T^{2} = {Kr}^{3}$
Here, $K$ is constant.
If masses of the sun and the planet are $M$ and $m$ respectively, then as per Newton's law and $m$ respectively, force of attraction between them is $F = \frac{G M m}{r^{2}}$ , where $G$ is gravitational constant.
The relation between $G$ and $K$ is described as

1

GK = 4 π^{2}

2

GMK = 4 π^{2}

3

K = G

4

K = \frac{1}{G}

Explanation:

B Gravitational force of attraction between sun and planet provides centripetal force for the orbit of planet.
$∴ \frac{GMm}{r^{2}} = \frac{{mv}^{2}}{r} =$ Centripetal force
$v^{2} = \frac{GM}{r} \dots \dots$
Time period of the planet is given by
$T = \frac{2 π r}{v}$
Now, squaring both side
$T^{2} = \frac{4 π^{2} r^{2}}{v^{2}}$
$T^{2} = \frac{4 π^{2} r^{2}}{(\frac{GM}{r})}$
$T^{2} = \frac{4 π^{2} r^{3}}{GM} \dots \dots$
According to question
$T^{2} = {Kr}^{3}$
Comparing equation (ii) and (iii), we get
${Kr}^{3} = \frac{4 π^{2} r^{3}}{GM}$
$K = \frac{4 π^{2}}{GM}$ $$
$∴ GMK = 4 π^{2}$

BCECE-2017

Gravitation

138610 A planet moving around the sun sweeps area $A_{1}$ in 2 days, $A_{2}$ in 4 days and $A_{3}$ in 9 days. Then, relation between them.

1

A_{1} = A_{2} = A_{3}

2

9 A_{1} = 3 A_{2} = 2 A_{3}

3

18 A_{1} = 9 A_{2} = 4 A_{3}

4

3 A_{1} = 4 A_{2} = 6 A_{3}

Explanation:

C Given that,
$A_{1} = 2 days$
$A_{2} = 4 days$
$A_{3} = 9 days$
According to the Kepler's second law,
$\frac{dA}{dt} = \frac{L}{2 m} = constant$
Where, $L =$ Angular momentum of the planet
$m =$ Mass of the planet
$\frac{dA}{dt} = constant$
$\frac{A_{1}}{2} = constant$
$\frac{A_{2}}{4} = constant$
$\frac{A_{3}}{9} = constant$
According to question-
$\frac{A_{1}}{2} = \frac{A_{2}}{4} = \frac{A_{3}}{9}$
Now we taking LCM 2, 4, 9 is equal to 36
Therefore,
$36 \times \frac{A_{1}}{2} = 36 \times \frac{A_{2}}{4} = 36 \times \frac{A_{3}}{9}$
$18 A_{1} = 9 A_{2} = 4 A_{3}$

BCECE-2018

Gravitation

138611 Two planets revolves around the sun with frequencies $N_{1}$ and $N_{2}$ revolutions per year. If their average radii (orbital) be $R_{1}$ and $R_{2}$ respectively, then $R_{1} / R_{2}$ is equal to-

1

{(N_{1} / N_{2})}^{2 / 3}

2

{(N_{1} / N_{2})}^{3 / 2}

3

{(N_{2} / N_{1})}^{2 / 3}

4

{(N_{2} / N_{1})}^{3 / 2}

Explanation:

C
According to Kepler's law of planetary motion $T^{2} \propto R^{3}$
According to question -
$T_{1} = \frac{1}{N_{1}}, T_{2} = \frac{1}{N_{2}}$
$∴ \frac{T^{2}}{R^{3}} =$ constant
$\frac{T_{1}^{2}}{R_{1}^{3}} = \frac{T_{2}^{2}}{R_{2}^{3}}$
$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{R_{1}^{3}}{R_{2}^{3}}$
$\frac{R_{1}}{R_{2}} = {(\frac{T_{1}}{T_{2}})}^{2 / 3}$
$∴ \frac{R_{1}}{R_{2}} = {(\frac{N_{2}}{N_{1}})}^{2 / 3}$

BCECE-2012

Gravitation

138614 The earth takes $24 hr$ to rotate once about its axis. How much time does the sun take to shift by $5^{\circ}$ when viewed from the earth?

1

20 \min

2

15 \min

3

10 \min

4

5 \min

Explanation:

A The Earth takes 24 hours to complete one rotation around own axis and complete one revolution about sun in 365 days.
A complete rotation of earth makes $360^{\circ} = 24 h$
$1^{\circ} = \frac{24 h}{360}$
$1^{\circ} = \frac{24 \times 60 \min}{360}$
$1^{\circ} = 4 \min$
$∴$ For $5^{\circ} = 5 \times 4 \min$
$= 20 \min$

J and K-CET-2014

Gravitation

138606 The distance of Neptune and Saturn from the sun is nearly $10^{13}$ and $10^{12}$ meter respectively. Assuming that they move in circular orbits, their periodic times will be in the ratio of

1 10

2 100

3

10 \sqrt{10}

4 1000

Explanation:

C Given that,

Now we using Kepler's third law,
$T^{2} \propto a^{3}$
$\frac{T_{N}^{2}}{T_{S}^{2}} = \frac{a_{N}^{3}}{a_{S}^{3}}$
${(\frac{T_{N}}{T_{S}})}^{2} = {(\frac{10^{13}}{10^{12}})}^{3} = 10^{3}$
$\frac{T_{N}}{T_{S}} = 10 \sqrt{10}$

AIPMT 1994

Gravitation

138609 Kepler's third law states that the square of period of revolution ( $T$ ) of a planet around the sun, is proportional to third power of average distance, $r$ between the sun and the planet i.e. $T^{2} = {Kr}^{3}$
Here, $K$ is constant.
If masses of the sun and the planet are $M$ and $m$ respectively, then as per Newton's law and $m$ respectively, force of attraction between them is $F = \frac{G M m}{r^{2}}$ , where $G$ is gravitational constant.
The relation between $G$ and $K$ is described as

1

GK = 4 π^{2}

2

GMK = 4 π^{2}

3

K = G

4

K = \frac{1}{G}

Explanation:

B Gravitational force of attraction between sun and planet provides centripetal force for the orbit of planet.
$∴ \frac{GMm}{r^{2}} = \frac{{mv}^{2}}{r} =$ Centripetal force
$v^{2} = \frac{GM}{r} \dots \dots$
Time period of the planet is given by
$T = \frac{2 π r}{v}$
Now, squaring both side
$T^{2} = \frac{4 π^{2} r^{2}}{v^{2}}$
$T^{2} = \frac{4 π^{2} r^{2}}{(\frac{GM}{r})}$
$T^{2} = \frac{4 π^{2} r^{3}}{GM} \dots \dots$
According to question
$T^{2} = {Kr}^{3}$
Comparing equation (ii) and (iii), we get
${Kr}^{3} = \frac{4 π^{2} r^{3}}{GM}$
$K = \frac{4 π^{2}}{GM}$ $$
$∴ GMK = 4 π^{2}$

BCECE-2017

Gravitation

138610 A planet moving around the sun sweeps area $A_{1}$ in 2 days, $A_{2}$ in 4 days and $A_{3}$ in 9 days. Then, relation between them.

1

A_{1} = A_{2} = A_{3}

2

9 A_{1} = 3 A_{2} = 2 A_{3}

3

18 A_{1} = 9 A_{2} = 4 A_{3}

4

3 A_{1} = 4 A_{2} = 6 A_{3}

Explanation:

C Given that,
$A_{1} = 2 days$
$A_{2} = 4 days$
$A_{3} = 9 days$
According to the Kepler's second law,
$\frac{dA}{dt} = \frac{L}{2 m} = constant$
Where, $L =$ Angular momentum of the planet
$m =$ Mass of the planet
$\frac{dA}{dt} = constant$
$\frac{A_{1}}{2} = constant$
$\frac{A_{2}}{4} = constant$
$\frac{A_{3}}{9} = constant$
According to question-
$\frac{A_{1}}{2} = \frac{A_{2}}{4} = \frac{A_{3}}{9}$
Now we taking LCM 2, 4, 9 is equal to 36
Therefore,
$36 \times \frac{A_{1}}{2} = 36 \times \frac{A_{2}}{4} = 36 \times \frac{A_{3}}{9}$
$18 A_{1} = 9 A_{2} = 4 A_{3}$

BCECE-2018

Gravitation

138611 Two planets revolves around the sun with frequencies $N_{1}$ and $N_{2}$ revolutions per year. If their average radii (orbital) be $R_{1}$ and $R_{2}$ respectively, then $R_{1} / R_{2}$ is equal to-

1

{(N_{1} / N_{2})}^{2 / 3}

2

{(N_{1} / N_{2})}^{3 / 2}

3

{(N_{2} / N_{1})}^{2 / 3}

4

{(N_{2} / N_{1})}^{3 / 2}

Explanation:

C
According to Kepler's law of planetary motion $T^{2} \propto R^{3}$
According to question -
$T_{1} = \frac{1}{N_{1}}, T_{2} = \frac{1}{N_{2}}$
$∴ \frac{T^{2}}{R^{3}} =$ constant
$\frac{T_{1}^{2}}{R_{1}^{3}} = \frac{T_{2}^{2}}{R_{2}^{3}}$
$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{R_{1}^{3}}{R_{2}^{3}}$
$\frac{R_{1}}{R_{2}} = {(\frac{T_{1}}{T_{2}})}^{2 / 3}$
$∴ \frac{R_{1}}{R_{2}} = {(\frac{N_{2}}{N_{1}})}^{2 / 3}$

BCECE-2012

Gravitation

138614 The earth takes $24 hr$ to rotate once about its axis. How much time does the sun take to shift by $5^{\circ}$ when viewed from the earth?

1

20 \min

2

15 \min

3

10 \min

4

5 \min

Explanation:

A The Earth takes 24 hours to complete one rotation around own axis and complete one revolution about sun in 365 days.
A complete rotation of earth makes $360^{\circ} = 24 h$
$1^{\circ} = \frac{24 h}{360}$
$1^{\circ} = \frac{24 \times 60 \min}{360}$
$1^{\circ} = 4 \min$
$∴$ For $5^{\circ} = 5 \times 4 \min$
$= 20 \min$

J and K-CET-2014