138592
Every planet revolves around the sun in an elliptical orbit:-
1 A and D only
2 B and C only
3 A and C only
4 C and D only
Explanation:
A According to Newton's gravitational law, $\mathrm{F}=\frac{\mathrm{GM}_{\mathrm{s}} \mathrm{m}_{\mathrm{p}}}{\mathrm{R}^{2}}$ $\mathrm{~F} \propto \frac{1}{\mathrm{r}^{2}}$ $\mathrm{~F} \propto \mathrm{m}_{1} \cdot \mathrm{m}_{2}$ This force provides centripetal force and acts towards sun. According to Kepler's third law - $\mathrm{T}^{2} \propto \mathrm{a}^{3}$
JEE Main-25.01.2023
Gravitation
138593
If the distance of the earth from Sun is $1.5 \times 10^{6}$ $\mathrm{km}$. Then the distance of an imaginary planet from Sun, if its period of revolution is $\mathbf{2 . 8 3}$ years is:
1 $6 \times 10^{6} \mathrm{~km}$
2 $3 \times 10^{6} \mathrm{~km}$
3 $6 \times 10^{7} \mathrm{~km}$
4 $3 \times 10^{7} \mathrm{~km}$
Explanation:
B Given that, Distance from Sun to Earth $\left(\mathrm{R}_{1}\right)=1.5 \times 10^{6} \mathrm{~m}$ Time period of planet $\left(T_{\text {planet }}\right)=2.83$ year Distance of an imaginary planet from $\operatorname{Sun}\left(\mathrm{R}_{2}\right)=$ ? We know that, $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\frac{\mathrm{R}_{1}^{3}}{\mathrm{R}_{2}^{3}}$ $\left(\frac{1}{2.83}\right)^{2}=\left(\frac{1.5 \times 10^{6}}{\mathrm{R}_{2}}\right)^{3}$ $\mathrm{R}_{2}^{3}=(2.83)^{2} \times\left(1.5 \times 10^{6}\right)^{3}$ $\mathrm{R}_{2}=3001112.10$ $\mathrm{R}_{2} \simeq 3 \times 10^{6} \mathrm{~km}$
JEE Main-24.01.2023
Gravitation
138594
The semi-major axis of the orbit of Saturn is approximately nine times that of Earth. The time period of revolution of Saturn is approximately equal to
1 81 years
2 27 years
3 729 years
4 $\sqrt[3]{81}$ years
5 9 years
Explanation:
B Given that, The semi-major axis of the orbit of Saturn is approximately nine times of Earth. Let $\left(r_{s}\right)$ and $\left(r_{e}\right)$ are semi-major axis of Saturn and Earth respectively. Now, using Kepler's thirds law $\mathrm{T}^{2} \propto \mathrm{r}^{3} \quad[\mathrm{~T}=\text { Time Period }]$ Then, $\quad T_{s}^{2} \propto r_{s}^{3}$ And $T_{e}^{2} \propto r_{e}^{3}$ Dividing equation (1) by equation (2) we get- $\frac{\mathrm{T}_{\mathrm{s}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}=\frac{\mathrm{r}_{\mathrm{s}}^{3}}{\mathrm{r}_{\mathrm{e}}^{3}}$ $\frac{\mathrm{T}_{\mathrm{s}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}=\left(\frac{\mathrm{r}_{\mathrm{s}}}{\mathrm{r}_{\mathrm{e}}}\right)^{3}$ According to question- $\because \quad r_{\mathrm{s}}=9 \mathrm{r}_{\mathrm{e}}$ Then, $\frac{\mathrm{T}_{\mathrm{s}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}=\left(\frac{9 \mathrm{r}_{\mathrm{e}}}{\mathrm{r}_{\mathrm{e}}}\right)^{3}$ $\frac{\mathrm{T}_{\mathrm{s}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}=(9)^{3}$ $\mathrm{T}_{\mathrm{s}}^{2}=\mathrm{T}_{\mathrm{e}}^{2}(9)^{3}$ $\mathrm{T}_{\mathrm{s}}=\sqrt{\mathrm{T}_{\mathrm{e}}^{2}(9)^{3}}$ $\mathrm{T}_{\mathrm{s}}=(9)^{3 / 2} \times \mathrm{T}_{\mathrm{e}}$ $\mathrm{T}_{\mathrm{s}}=\left(3^{2}\right)^{3 / 2} \times 1$ Year $\quad\left[\mathrm{T}_{\mathrm{e}}=1\right.$ year $]$ $\mathrm{T}_{\mathrm{s}}=(3)^{3}$ $\mathrm{T}_{\mathrm{s}}=27$ years
Kerala CEE - 2017
Gravitation
138595
A comet orbits around the Sun in an elliptical orbit. Which of the following quantities remains constant during the course of its motion?
1 Linear velocity
2 Angular velocity
3 Angular momentum
4 Kinetic energy
5 Potential energy
Explanation:
C Kepler's second law- Comet and sun moves in elliptical orbit equal area swept in equal interval of time. $\frac{\mathrm{L}}{2 \mathrm{M}}=$ Constant (since no external torque) $\mathrm{L}=$ constant The angular momentum of comet is same throughout due to conservation of angular momentum in absence of any torque.
Kerala CEE - 2017
Gravitation
138596
If the earth is one-fourth of its present distance from the sun, the duration of the year will be changed to
1 half of the present year
2 $\frac{1}{4}$ th of the present year
3 $\frac{1}{8}$ th of the present year
4 $\frac{7}{8}$ th of the present year
5 $\frac{1}{16}$ th of the present year
Explanation:
C Given that, If the earth is one-fourth of its present distance from the sun. $\therefore \quad \mathrm{r}_{2}=\frac{1}{4} \mathrm{r}_{1}$ By Kepler's $3^{\text {rd }}$ law- $\mathrm{T}^{2} \propto \mathrm{r}^{3}$ Then $\quad \mathrm{T}_{1}^{2} \propto \mathrm{r}_{1}^{3}$ and $\quad \mathrm{T}_{2}^{2} \propto \mathrm{r}_{2}^{3}$ Dividing equation (1) by equation (2) we get- $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\frac{\mathrm{r}_{1}^{3}}{\mathrm{r}_{2}^{3}}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\left(\frac{\mathrm{r}_{1}}{\frac{1}{4} \mathrm{r}_{1}}\right)^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=(4)^{3}$ $\mathrm{~T}_{1}^{2}=\mathrm{T}_{2}^{2}(4)^{3}$ $\mathrm{~T}_{2}^{2}=\frac{\mathrm{T}_{1}^{2}}{(4)^{3}}$ $\mathrm{~T}_{2}=\frac{\mathrm{T}_{1}}{4^{3 / 2}}$ $\mathrm{~T}_{2}=\frac{\mathrm{T}_{1}}{8}$
138592
Every planet revolves around the sun in an elliptical orbit:-
1 A and D only
2 B and C only
3 A and C only
4 C and D only
Explanation:
A According to Newton's gravitational law, $\mathrm{F}=\frac{\mathrm{GM}_{\mathrm{s}} \mathrm{m}_{\mathrm{p}}}{\mathrm{R}^{2}}$ $\mathrm{~F} \propto \frac{1}{\mathrm{r}^{2}}$ $\mathrm{~F} \propto \mathrm{m}_{1} \cdot \mathrm{m}_{2}$ This force provides centripetal force and acts towards sun. According to Kepler's third law - $\mathrm{T}^{2} \propto \mathrm{a}^{3}$
JEE Main-25.01.2023
Gravitation
138593
If the distance of the earth from Sun is $1.5 \times 10^{6}$ $\mathrm{km}$. Then the distance of an imaginary planet from Sun, if its period of revolution is $\mathbf{2 . 8 3}$ years is:
1 $6 \times 10^{6} \mathrm{~km}$
2 $3 \times 10^{6} \mathrm{~km}$
3 $6 \times 10^{7} \mathrm{~km}$
4 $3 \times 10^{7} \mathrm{~km}$
Explanation:
B Given that, Distance from Sun to Earth $\left(\mathrm{R}_{1}\right)=1.5 \times 10^{6} \mathrm{~m}$ Time period of planet $\left(T_{\text {planet }}\right)=2.83$ year Distance of an imaginary planet from $\operatorname{Sun}\left(\mathrm{R}_{2}\right)=$ ? We know that, $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\frac{\mathrm{R}_{1}^{3}}{\mathrm{R}_{2}^{3}}$ $\left(\frac{1}{2.83}\right)^{2}=\left(\frac{1.5 \times 10^{6}}{\mathrm{R}_{2}}\right)^{3}$ $\mathrm{R}_{2}^{3}=(2.83)^{2} \times\left(1.5 \times 10^{6}\right)^{3}$ $\mathrm{R}_{2}=3001112.10$ $\mathrm{R}_{2} \simeq 3 \times 10^{6} \mathrm{~km}$
JEE Main-24.01.2023
Gravitation
138594
The semi-major axis of the orbit of Saturn is approximately nine times that of Earth. The time period of revolution of Saturn is approximately equal to
1 81 years
2 27 years
3 729 years
4 $\sqrt[3]{81}$ years
5 9 years
Explanation:
B Given that, The semi-major axis of the orbit of Saturn is approximately nine times of Earth. Let $\left(r_{s}\right)$ and $\left(r_{e}\right)$ are semi-major axis of Saturn and Earth respectively. Now, using Kepler's thirds law $\mathrm{T}^{2} \propto \mathrm{r}^{3} \quad[\mathrm{~T}=\text { Time Period }]$ Then, $\quad T_{s}^{2} \propto r_{s}^{3}$ And $T_{e}^{2} \propto r_{e}^{3}$ Dividing equation (1) by equation (2) we get- $\frac{\mathrm{T}_{\mathrm{s}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}=\frac{\mathrm{r}_{\mathrm{s}}^{3}}{\mathrm{r}_{\mathrm{e}}^{3}}$ $\frac{\mathrm{T}_{\mathrm{s}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}=\left(\frac{\mathrm{r}_{\mathrm{s}}}{\mathrm{r}_{\mathrm{e}}}\right)^{3}$ According to question- $\because \quad r_{\mathrm{s}}=9 \mathrm{r}_{\mathrm{e}}$ Then, $\frac{\mathrm{T}_{\mathrm{s}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}=\left(\frac{9 \mathrm{r}_{\mathrm{e}}}{\mathrm{r}_{\mathrm{e}}}\right)^{3}$ $\frac{\mathrm{T}_{\mathrm{s}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}=(9)^{3}$ $\mathrm{T}_{\mathrm{s}}^{2}=\mathrm{T}_{\mathrm{e}}^{2}(9)^{3}$ $\mathrm{T}_{\mathrm{s}}=\sqrt{\mathrm{T}_{\mathrm{e}}^{2}(9)^{3}}$ $\mathrm{T}_{\mathrm{s}}=(9)^{3 / 2} \times \mathrm{T}_{\mathrm{e}}$ $\mathrm{T}_{\mathrm{s}}=\left(3^{2}\right)^{3 / 2} \times 1$ Year $\quad\left[\mathrm{T}_{\mathrm{e}}=1\right.$ year $]$ $\mathrm{T}_{\mathrm{s}}=(3)^{3}$ $\mathrm{T}_{\mathrm{s}}=27$ years
Kerala CEE - 2017
Gravitation
138595
A comet orbits around the Sun in an elliptical orbit. Which of the following quantities remains constant during the course of its motion?
1 Linear velocity
2 Angular velocity
3 Angular momentum
4 Kinetic energy
5 Potential energy
Explanation:
C Kepler's second law- Comet and sun moves in elliptical orbit equal area swept in equal interval of time. $\frac{\mathrm{L}}{2 \mathrm{M}}=$ Constant (since no external torque) $\mathrm{L}=$ constant The angular momentum of comet is same throughout due to conservation of angular momentum in absence of any torque.
Kerala CEE - 2017
Gravitation
138596
If the earth is one-fourth of its present distance from the sun, the duration of the year will be changed to
1 half of the present year
2 $\frac{1}{4}$ th of the present year
3 $\frac{1}{8}$ th of the present year
4 $\frac{7}{8}$ th of the present year
5 $\frac{1}{16}$ th of the present year
Explanation:
C Given that, If the earth is one-fourth of its present distance from the sun. $\therefore \quad \mathrm{r}_{2}=\frac{1}{4} \mathrm{r}_{1}$ By Kepler's $3^{\text {rd }}$ law- $\mathrm{T}^{2} \propto \mathrm{r}^{3}$ Then $\quad \mathrm{T}_{1}^{2} \propto \mathrm{r}_{1}^{3}$ and $\quad \mathrm{T}_{2}^{2} \propto \mathrm{r}_{2}^{3}$ Dividing equation (1) by equation (2) we get- $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\frac{\mathrm{r}_{1}^{3}}{\mathrm{r}_{2}^{3}}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\left(\frac{\mathrm{r}_{1}}{\frac{1}{4} \mathrm{r}_{1}}\right)^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=(4)^{3}$ $\mathrm{~T}_{1}^{2}=\mathrm{T}_{2}^{2}(4)^{3}$ $\mathrm{~T}_{2}^{2}=\frac{\mathrm{T}_{1}^{2}}{(4)^{3}}$ $\mathrm{~T}_{2}=\frac{\mathrm{T}_{1}}{4^{3 / 2}}$ $\mathrm{~T}_{2}=\frac{\mathrm{T}_{1}}{8}$
138592
Every planet revolves around the sun in an elliptical orbit:-
1 A and D only
2 B and C only
3 A and C only
4 C and D only
Explanation:
A According to Newton's gravitational law, $\mathrm{F}=\frac{\mathrm{GM}_{\mathrm{s}} \mathrm{m}_{\mathrm{p}}}{\mathrm{R}^{2}}$ $\mathrm{~F} \propto \frac{1}{\mathrm{r}^{2}}$ $\mathrm{~F} \propto \mathrm{m}_{1} \cdot \mathrm{m}_{2}$ This force provides centripetal force and acts towards sun. According to Kepler's third law - $\mathrm{T}^{2} \propto \mathrm{a}^{3}$
JEE Main-25.01.2023
Gravitation
138593
If the distance of the earth from Sun is $1.5 \times 10^{6}$ $\mathrm{km}$. Then the distance of an imaginary planet from Sun, if its period of revolution is $\mathbf{2 . 8 3}$ years is:
1 $6 \times 10^{6} \mathrm{~km}$
2 $3 \times 10^{6} \mathrm{~km}$
3 $6 \times 10^{7} \mathrm{~km}$
4 $3 \times 10^{7} \mathrm{~km}$
Explanation:
B Given that, Distance from Sun to Earth $\left(\mathrm{R}_{1}\right)=1.5 \times 10^{6} \mathrm{~m}$ Time period of planet $\left(T_{\text {planet }}\right)=2.83$ year Distance of an imaginary planet from $\operatorname{Sun}\left(\mathrm{R}_{2}\right)=$ ? We know that, $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\frac{\mathrm{R}_{1}^{3}}{\mathrm{R}_{2}^{3}}$ $\left(\frac{1}{2.83}\right)^{2}=\left(\frac{1.5 \times 10^{6}}{\mathrm{R}_{2}}\right)^{3}$ $\mathrm{R}_{2}^{3}=(2.83)^{2} \times\left(1.5 \times 10^{6}\right)^{3}$ $\mathrm{R}_{2}=3001112.10$ $\mathrm{R}_{2} \simeq 3 \times 10^{6} \mathrm{~km}$
JEE Main-24.01.2023
Gravitation
138594
The semi-major axis of the orbit of Saturn is approximately nine times that of Earth. The time period of revolution of Saturn is approximately equal to
1 81 years
2 27 years
3 729 years
4 $\sqrt[3]{81}$ years
5 9 years
Explanation:
B Given that, The semi-major axis of the orbit of Saturn is approximately nine times of Earth. Let $\left(r_{s}\right)$ and $\left(r_{e}\right)$ are semi-major axis of Saturn and Earth respectively. Now, using Kepler's thirds law $\mathrm{T}^{2} \propto \mathrm{r}^{3} \quad[\mathrm{~T}=\text { Time Period }]$ Then, $\quad T_{s}^{2} \propto r_{s}^{3}$ And $T_{e}^{2} \propto r_{e}^{3}$ Dividing equation (1) by equation (2) we get- $\frac{\mathrm{T}_{\mathrm{s}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}=\frac{\mathrm{r}_{\mathrm{s}}^{3}}{\mathrm{r}_{\mathrm{e}}^{3}}$ $\frac{\mathrm{T}_{\mathrm{s}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}=\left(\frac{\mathrm{r}_{\mathrm{s}}}{\mathrm{r}_{\mathrm{e}}}\right)^{3}$ According to question- $\because \quad r_{\mathrm{s}}=9 \mathrm{r}_{\mathrm{e}}$ Then, $\frac{\mathrm{T}_{\mathrm{s}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}=\left(\frac{9 \mathrm{r}_{\mathrm{e}}}{\mathrm{r}_{\mathrm{e}}}\right)^{3}$ $\frac{\mathrm{T}_{\mathrm{s}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}=(9)^{3}$ $\mathrm{T}_{\mathrm{s}}^{2}=\mathrm{T}_{\mathrm{e}}^{2}(9)^{3}$ $\mathrm{T}_{\mathrm{s}}=\sqrt{\mathrm{T}_{\mathrm{e}}^{2}(9)^{3}}$ $\mathrm{T}_{\mathrm{s}}=(9)^{3 / 2} \times \mathrm{T}_{\mathrm{e}}$ $\mathrm{T}_{\mathrm{s}}=\left(3^{2}\right)^{3 / 2} \times 1$ Year $\quad\left[\mathrm{T}_{\mathrm{e}}=1\right.$ year $]$ $\mathrm{T}_{\mathrm{s}}=(3)^{3}$ $\mathrm{T}_{\mathrm{s}}=27$ years
Kerala CEE - 2017
Gravitation
138595
A comet orbits around the Sun in an elliptical orbit. Which of the following quantities remains constant during the course of its motion?
1 Linear velocity
2 Angular velocity
3 Angular momentum
4 Kinetic energy
5 Potential energy
Explanation:
C Kepler's second law- Comet and sun moves in elliptical orbit equal area swept in equal interval of time. $\frac{\mathrm{L}}{2 \mathrm{M}}=$ Constant (since no external torque) $\mathrm{L}=$ constant The angular momentum of comet is same throughout due to conservation of angular momentum in absence of any torque.
Kerala CEE - 2017
Gravitation
138596
If the earth is one-fourth of its present distance from the sun, the duration of the year will be changed to
1 half of the present year
2 $\frac{1}{4}$ th of the present year
3 $\frac{1}{8}$ th of the present year
4 $\frac{7}{8}$ th of the present year
5 $\frac{1}{16}$ th of the present year
Explanation:
C Given that, If the earth is one-fourth of its present distance from the sun. $\therefore \quad \mathrm{r}_{2}=\frac{1}{4} \mathrm{r}_{1}$ By Kepler's $3^{\text {rd }}$ law- $\mathrm{T}^{2} \propto \mathrm{r}^{3}$ Then $\quad \mathrm{T}_{1}^{2} \propto \mathrm{r}_{1}^{3}$ and $\quad \mathrm{T}_{2}^{2} \propto \mathrm{r}_{2}^{3}$ Dividing equation (1) by equation (2) we get- $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\frac{\mathrm{r}_{1}^{3}}{\mathrm{r}_{2}^{3}}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\left(\frac{\mathrm{r}_{1}}{\frac{1}{4} \mathrm{r}_{1}}\right)^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=(4)^{3}$ $\mathrm{~T}_{1}^{2}=\mathrm{T}_{2}^{2}(4)^{3}$ $\mathrm{~T}_{2}^{2}=\frac{\mathrm{T}_{1}^{2}}{(4)^{3}}$ $\mathrm{~T}_{2}=\frac{\mathrm{T}_{1}}{4^{3 / 2}}$ $\mathrm{~T}_{2}=\frac{\mathrm{T}_{1}}{8}$
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Gravitation
138592
Every planet revolves around the sun in an elliptical orbit:-
1 A and D only
2 B and C only
3 A and C only
4 C and D only
Explanation:
A According to Newton's gravitational law, $\mathrm{F}=\frac{\mathrm{GM}_{\mathrm{s}} \mathrm{m}_{\mathrm{p}}}{\mathrm{R}^{2}}$ $\mathrm{~F} \propto \frac{1}{\mathrm{r}^{2}}$ $\mathrm{~F} \propto \mathrm{m}_{1} \cdot \mathrm{m}_{2}$ This force provides centripetal force and acts towards sun. According to Kepler's third law - $\mathrm{T}^{2} \propto \mathrm{a}^{3}$
JEE Main-25.01.2023
Gravitation
138593
If the distance of the earth from Sun is $1.5 \times 10^{6}$ $\mathrm{km}$. Then the distance of an imaginary planet from Sun, if its period of revolution is $\mathbf{2 . 8 3}$ years is:
1 $6 \times 10^{6} \mathrm{~km}$
2 $3 \times 10^{6} \mathrm{~km}$
3 $6 \times 10^{7} \mathrm{~km}$
4 $3 \times 10^{7} \mathrm{~km}$
Explanation:
B Given that, Distance from Sun to Earth $\left(\mathrm{R}_{1}\right)=1.5 \times 10^{6} \mathrm{~m}$ Time period of planet $\left(T_{\text {planet }}\right)=2.83$ year Distance of an imaginary planet from $\operatorname{Sun}\left(\mathrm{R}_{2}\right)=$ ? We know that, $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\frac{\mathrm{R}_{1}^{3}}{\mathrm{R}_{2}^{3}}$ $\left(\frac{1}{2.83}\right)^{2}=\left(\frac{1.5 \times 10^{6}}{\mathrm{R}_{2}}\right)^{3}$ $\mathrm{R}_{2}^{3}=(2.83)^{2} \times\left(1.5 \times 10^{6}\right)^{3}$ $\mathrm{R}_{2}=3001112.10$ $\mathrm{R}_{2} \simeq 3 \times 10^{6} \mathrm{~km}$
JEE Main-24.01.2023
Gravitation
138594
The semi-major axis of the orbit of Saturn is approximately nine times that of Earth. The time period of revolution of Saturn is approximately equal to
1 81 years
2 27 years
3 729 years
4 $\sqrt[3]{81}$ years
5 9 years
Explanation:
B Given that, The semi-major axis of the orbit of Saturn is approximately nine times of Earth. Let $\left(r_{s}\right)$ and $\left(r_{e}\right)$ are semi-major axis of Saturn and Earth respectively. Now, using Kepler's thirds law $\mathrm{T}^{2} \propto \mathrm{r}^{3} \quad[\mathrm{~T}=\text { Time Period }]$ Then, $\quad T_{s}^{2} \propto r_{s}^{3}$ And $T_{e}^{2} \propto r_{e}^{3}$ Dividing equation (1) by equation (2) we get- $\frac{\mathrm{T}_{\mathrm{s}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}=\frac{\mathrm{r}_{\mathrm{s}}^{3}}{\mathrm{r}_{\mathrm{e}}^{3}}$ $\frac{\mathrm{T}_{\mathrm{s}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}=\left(\frac{\mathrm{r}_{\mathrm{s}}}{\mathrm{r}_{\mathrm{e}}}\right)^{3}$ According to question- $\because \quad r_{\mathrm{s}}=9 \mathrm{r}_{\mathrm{e}}$ Then, $\frac{\mathrm{T}_{\mathrm{s}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}=\left(\frac{9 \mathrm{r}_{\mathrm{e}}}{\mathrm{r}_{\mathrm{e}}}\right)^{3}$ $\frac{\mathrm{T}_{\mathrm{s}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}=(9)^{3}$ $\mathrm{T}_{\mathrm{s}}^{2}=\mathrm{T}_{\mathrm{e}}^{2}(9)^{3}$ $\mathrm{T}_{\mathrm{s}}=\sqrt{\mathrm{T}_{\mathrm{e}}^{2}(9)^{3}}$ $\mathrm{T}_{\mathrm{s}}=(9)^{3 / 2} \times \mathrm{T}_{\mathrm{e}}$ $\mathrm{T}_{\mathrm{s}}=\left(3^{2}\right)^{3 / 2} \times 1$ Year $\quad\left[\mathrm{T}_{\mathrm{e}}=1\right.$ year $]$ $\mathrm{T}_{\mathrm{s}}=(3)^{3}$ $\mathrm{T}_{\mathrm{s}}=27$ years
Kerala CEE - 2017
Gravitation
138595
A comet orbits around the Sun in an elliptical orbit. Which of the following quantities remains constant during the course of its motion?
1 Linear velocity
2 Angular velocity
3 Angular momentum
4 Kinetic energy
5 Potential energy
Explanation:
C Kepler's second law- Comet and sun moves in elliptical orbit equal area swept in equal interval of time. $\frac{\mathrm{L}}{2 \mathrm{M}}=$ Constant (since no external torque) $\mathrm{L}=$ constant The angular momentum of comet is same throughout due to conservation of angular momentum in absence of any torque.
Kerala CEE - 2017
Gravitation
138596
If the earth is one-fourth of its present distance from the sun, the duration of the year will be changed to
1 half of the present year
2 $\frac{1}{4}$ th of the present year
3 $\frac{1}{8}$ th of the present year
4 $\frac{7}{8}$ th of the present year
5 $\frac{1}{16}$ th of the present year
Explanation:
C Given that, If the earth is one-fourth of its present distance from the sun. $\therefore \quad \mathrm{r}_{2}=\frac{1}{4} \mathrm{r}_{1}$ By Kepler's $3^{\text {rd }}$ law- $\mathrm{T}^{2} \propto \mathrm{r}^{3}$ Then $\quad \mathrm{T}_{1}^{2} \propto \mathrm{r}_{1}^{3}$ and $\quad \mathrm{T}_{2}^{2} \propto \mathrm{r}_{2}^{3}$ Dividing equation (1) by equation (2) we get- $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\frac{\mathrm{r}_{1}^{3}}{\mathrm{r}_{2}^{3}}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\left(\frac{\mathrm{r}_{1}}{\frac{1}{4} \mathrm{r}_{1}}\right)^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=(4)^{3}$ $\mathrm{~T}_{1}^{2}=\mathrm{T}_{2}^{2}(4)^{3}$ $\mathrm{~T}_{2}^{2}=\frac{\mathrm{T}_{1}^{2}}{(4)^{3}}$ $\mathrm{~T}_{2}=\frac{\mathrm{T}_{1}}{4^{3 / 2}}$ $\mathrm{~T}_{2}=\frac{\mathrm{T}_{1}}{8}$
138592
Every planet revolves around the sun in an elliptical orbit:-
1 A and D only
2 B and C only
3 A and C only
4 C and D only
Explanation:
A According to Newton's gravitational law, $\mathrm{F}=\frac{\mathrm{GM}_{\mathrm{s}} \mathrm{m}_{\mathrm{p}}}{\mathrm{R}^{2}}$ $\mathrm{~F} \propto \frac{1}{\mathrm{r}^{2}}$ $\mathrm{~F} \propto \mathrm{m}_{1} \cdot \mathrm{m}_{2}$ This force provides centripetal force and acts towards sun. According to Kepler's third law - $\mathrm{T}^{2} \propto \mathrm{a}^{3}$
JEE Main-25.01.2023
Gravitation
138593
If the distance of the earth from Sun is $1.5 \times 10^{6}$ $\mathrm{km}$. Then the distance of an imaginary planet from Sun, if its period of revolution is $\mathbf{2 . 8 3}$ years is:
1 $6 \times 10^{6} \mathrm{~km}$
2 $3 \times 10^{6} \mathrm{~km}$
3 $6 \times 10^{7} \mathrm{~km}$
4 $3 \times 10^{7} \mathrm{~km}$
Explanation:
B Given that, Distance from Sun to Earth $\left(\mathrm{R}_{1}\right)=1.5 \times 10^{6} \mathrm{~m}$ Time period of planet $\left(T_{\text {planet }}\right)=2.83$ year Distance of an imaginary planet from $\operatorname{Sun}\left(\mathrm{R}_{2}\right)=$ ? We know that, $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\frac{\mathrm{R}_{1}^{3}}{\mathrm{R}_{2}^{3}}$ $\left(\frac{1}{2.83}\right)^{2}=\left(\frac{1.5 \times 10^{6}}{\mathrm{R}_{2}}\right)^{3}$ $\mathrm{R}_{2}^{3}=(2.83)^{2} \times\left(1.5 \times 10^{6}\right)^{3}$ $\mathrm{R}_{2}=3001112.10$ $\mathrm{R}_{2} \simeq 3 \times 10^{6} \mathrm{~km}$
JEE Main-24.01.2023
Gravitation
138594
The semi-major axis of the orbit of Saturn is approximately nine times that of Earth. The time period of revolution of Saturn is approximately equal to
1 81 years
2 27 years
3 729 years
4 $\sqrt[3]{81}$ years
5 9 years
Explanation:
B Given that, The semi-major axis of the orbit of Saturn is approximately nine times of Earth. Let $\left(r_{s}\right)$ and $\left(r_{e}\right)$ are semi-major axis of Saturn and Earth respectively. Now, using Kepler's thirds law $\mathrm{T}^{2} \propto \mathrm{r}^{3} \quad[\mathrm{~T}=\text { Time Period }]$ Then, $\quad T_{s}^{2} \propto r_{s}^{3}$ And $T_{e}^{2} \propto r_{e}^{3}$ Dividing equation (1) by equation (2) we get- $\frac{\mathrm{T}_{\mathrm{s}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}=\frac{\mathrm{r}_{\mathrm{s}}^{3}}{\mathrm{r}_{\mathrm{e}}^{3}}$ $\frac{\mathrm{T}_{\mathrm{s}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}=\left(\frac{\mathrm{r}_{\mathrm{s}}}{\mathrm{r}_{\mathrm{e}}}\right)^{3}$ According to question- $\because \quad r_{\mathrm{s}}=9 \mathrm{r}_{\mathrm{e}}$ Then, $\frac{\mathrm{T}_{\mathrm{s}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}=\left(\frac{9 \mathrm{r}_{\mathrm{e}}}{\mathrm{r}_{\mathrm{e}}}\right)^{3}$ $\frac{\mathrm{T}_{\mathrm{s}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}=(9)^{3}$ $\mathrm{T}_{\mathrm{s}}^{2}=\mathrm{T}_{\mathrm{e}}^{2}(9)^{3}$ $\mathrm{T}_{\mathrm{s}}=\sqrt{\mathrm{T}_{\mathrm{e}}^{2}(9)^{3}}$ $\mathrm{T}_{\mathrm{s}}=(9)^{3 / 2} \times \mathrm{T}_{\mathrm{e}}$ $\mathrm{T}_{\mathrm{s}}=\left(3^{2}\right)^{3 / 2} \times 1$ Year $\quad\left[\mathrm{T}_{\mathrm{e}}=1\right.$ year $]$ $\mathrm{T}_{\mathrm{s}}=(3)^{3}$ $\mathrm{T}_{\mathrm{s}}=27$ years
Kerala CEE - 2017
Gravitation
138595
A comet orbits around the Sun in an elliptical orbit. Which of the following quantities remains constant during the course of its motion?
1 Linear velocity
2 Angular velocity
3 Angular momentum
4 Kinetic energy
5 Potential energy
Explanation:
C Kepler's second law- Comet and sun moves in elliptical orbit equal area swept in equal interval of time. $\frac{\mathrm{L}}{2 \mathrm{M}}=$ Constant (since no external torque) $\mathrm{L}=$ constant The angular momentum of comet is same throughout due to conservation of angular momentum in absence of any torque.
Kerala CEE - 2017
Gravitation
138596
If the earth is one-fourth of its present distance from the sun, the duration of the year will be changed to
1 half of the present year
2 $\frac{1}{4}$ th of the present year
3 $\frac{1}{8}$ th of the present year
4 $\frac{7}{8}$ th of the present year
5 $\frac{1}{16}$ th of the present year
Explanation:
C Given that, If the earth is one-fourth of its present distance from the sun. $\therefore \quad \mathrm{r}_{2}=\frac{1}{4} \mathrm{r}_{1}$ By Kepler's $3^{\text {rd }}$ law- $\mathrm{T}^{2} \propto \mathrm{r}^{3}$ Then $\quad \mathrm{T}_{1}^{2} \propto \mathrm{r}_{1}^{3}$ and $\quad \mathrm{T}_{2}^{2} \propto \mathrm{r}_{2}^{3}$ Dividing equation (1) by equation (2) we get- $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\frac{\mathrm{r}_{1}^{3}}{\mathrm{r}_{2}^{3}}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\left(\frac{\mathrm{r}_{1}}{\frac{1}{4} \mathrm{r}_{1}}\right)^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=(4)^{3}$ $\mathrm{~T}_{1}^{2}=\mathrm{T}_{2}^{2}(4)^{3}$ $\mathrm{~T}_{2}^{2}=\frac{\mathrm{T}_{1}^{2}}{(4)^{3}}$ $\mathrm{~T}_{2}=\frac{\mathrm{T}_{1}}{4^{3 / 2}}$ $\mathrm{~T}_{2}=\frac{\mathrm{T}_{1}}{8}$
