NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Gravitation
138650
Assume a planet with orbiting radius $R$ and period of revolution $T$ around the sun experiences a gravitational force which follows inverse cube law instead of inverse square law. In this case, the period of revolution is proportional to
1 $\frac{1}{\mathrm{R}}$
2 $\mathrm{R}^{2}$
3 $\mathrm{R}^{3 / 2}$
4 $\mathrm{R}^{4}$
Explanation:
B If the law of gravitation becomes an inverse cube law instead of inverse square law, $\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{R}^{3}}$ For a planet of mass ' $\mathrm{m}$ ' revolving around the sun, we can write $\frac{\mathrm{GMm}}{\mathrm{R}^{3}}=\frac{\mathrm{mv}^{2}}{\mathrm{R}}$ $\mathrm{v}=\frac{\sqrt{\mathrm{GM}}}{\mathrm{R}}$ Time period of revolution of planet $\mathrm{T} =\frac{2 \pi R}{\mathrm{~V}}$ $\mathrm{~T} =\frac{2 \pi \mathrm{R}}{\frac{\sqrt{\mathrm{Gm}}}{\mathrm{R}}}$ $\mathrm{T} =\frac{2 \pi \mathrm{R}^{2}}{\sqrt{\mathrm{Gm}}}$ $\mathrm{T} \propto \mathrm{R}^{2}$
AP EAMCET-11.07.2022
Gravitation
138651
The period of revolution of an early satellite close to the surface of earth is 90 minutes. The time period of another satellite in an orbit at a distance of three times the radius of earth from the earth's surface will be
1 $90 \sqrt{8}$ minutes
2 360 minutes
3 720 minutes
4 270 minutes
Explanation:
C Given, $\mathrm{R}_{1}=\mathrm{R}_{\mathrm{e}}$ $\mathrm{R}_{2}=\mathrm{R}_{\mathrm{e}}+3 \mathrm{R}_{\mathrm{e}}=4 \mathrm{R}_{\mathrm{e}}$ $\mathrm{T}_{1}=90 \text { min. }$ From, Kepler's third law of planetary motion, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ So, $\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{2}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3}$ $\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{2}=\left(\frac{\mathrm{R}_{1}}{4 \mathrm{R}_{\mathrm{e}}}\right)^{3}$ $\frac{(90)^{2}}{\left(\mathrm{~T}_{2}\right)^{2}}=\frac{1}{64}$ $\mathrm{~T}_{2}=\sqrt{(90)^{2} \times 64}$ $\mathrm{~T}_{2}=90 \times 8$ $\mathrm{~T}_{2}=720$
AP EAMCET-28.04.2017
Gravitation
138652
A small satellite is in elliptical orbit around the earth as shown. $L$ denotes the magnitude of its angular momentum and $K$ denotes its kinetic energy. 1 and 2 denote two positions of the satellite, then
B By kepler's law, when a satellite is moving around the earth on elliptical path, then its angular momentum remains constant. $\mathrm{L}=\operatorname{mvr} \text { (angular momentum) }$ $\mathrm{K}=\frac{1}{2} \mathrm{mv}^2 \text { (Kinetic energy) }$ $\because \quad \mathrm{L}_1=\mathrm{L}_2$ $\mathrm{~m}_1 \mathrm{v}_1 \mathrm{r}_1=\mathrm{m}_2 \mathrm{v}_2 \mathrm{r}_2$ $\text { But, } \quad \mathrm{m}_1=\mathrm{m}_2=\mathrm{m}$ $\therefore \quad \mathrm{v}_1 \mathrm{r}_1=\mathrm{v}_2 \mathrm{r}_2$ $\frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{\mathrm{v}_2}{\mathrm{v}_1}$ $\text { Here, } \quad \mathrm{r}_1>\mathrm{r}_2$ $\frac{\mathrm{r}_1}{\mathrm{r}_2}>1$ $\frac{\mathrm{v}_2}{\mathrm{v}_1}>1$ $\mathrm{v}_2>\mathrm{v}_1$ $\mathrm{v}_2^2>\mathrm{v}_1^2 \text { or } \frac{1}{2} \mathrm{mv}_2^2>\frac{1}{2} \mathrm{mv}_1^2$ $\mathrm{~K}_2>\mathrm{K}_1\left(\because \mathrm{K}=\frac{1}{2} \mathrm{mv}^2\right)$ From equation (i), \(\frac{\mathrm{v}_2}{\mathrm{v}_1}>1\) \(\mathrm{v}_2>\mathrm{v}_1\) \(\mathrm{v}_2^2>\mathrm{v}_1^2 \text { or } \frac{1}{2} \mathrm{mv}_2^2>\frac{1}{2} \mathrm{mv}_1^2\) \(\mathrm{~K}_2>\mathrm{K}_1\left(\because \mathrm{K}=\frac{1}{2} \mathrm{mv}^2\right)\)
AMU-2018
Gravitation
138653
A planet is moving in an elliptical orbit around the sun. If $T, U, E$ and $L$ stand for its kinetic energy, gravitational potential energy, total energy and magnitude of angular momentum about the centre of force, which of the following is correct?
1 $\mathrm{T}$ is conserved
2 $\mathrm{U}$ is always positive
3 $\mathrm{E}$ is always negative
4 $\mathrm{L}$ is conserved but direction of vector $\mathrm{L}$ changes continuously
Explanation:
C Angular momentum conserved with direction of $\overrightarrow{\mathrm{L}}$ always in one direction. $\overrightarrow{\mathrm{L}}=\mathrm{mvr}=\text { constant }$ $\text { K.E. }=\mathrm{T}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{U}=-\frac{\mathrm{GMm}}{\mathrm{r}}$ For the planet to move in any orbit, its total mechanical energy must be bound then $\mathrm{E}$ must be negative always. Kinetic energy (T):- The path of the planet is given to be elliptical, so the distance between the center and the planet will change. Due to this change, As a result, the kinetic energy will not be conserved. Potential energy (U): For any attractive field, the potential energy is negative. Since the work is done against the acceleration, hence there is always a negative sign potential energy.
138650
Assume a planet with orbiting radius $R$ and period of revolution $T$ around the sun experiences a gravitational force which follows inverse cube law instead of inverse square law. In this case, the period of revolution is proportional to
1 $\frac{1}{\mathrm{R}}$
2 $\mathrm{R}^{2}$
3 $\mathrm{R}^{3 / 2}$
4 $\mathrm{R}^{4}$
Explanation:
B If the law of gravitation becomes an inverse cube law instead of inverse square law, $\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{R}^{3}}$ For a planet of mass ' $\mathrm{m}$ ' revolving around the sun, we can write $\frac{\mathrm{GMm}}{\mathrm{R}^{3}}=\frac{\mathrm{mv}^{2}}{\mathrm{R}}$ $\mathrm{v}=\frac{\sqrt{\mathrm{GM}}}{\mathrm{R}}$ Time period of revolution of planet $\mathrm{T} =\frac{2 \pi R}{\mathrm{~V}}$ $\mathrm{~T} =\frac{2 \pi \mathrm{R}}{\frac{\sqrt{\mathrm{Gm}}}{\mathrm{R}}}$ $\mathrm{T} =\frac{2 \pi \mathrm{R}^{2}}{\sqrt{\mathrm{Gm}}}$ $\mathrm{T} \propto \mathrm{R}^{2}$
AP EAMCET-11.07.2022
Gravitation
138651
The period of revolution of an early satellite close to the surface of earth is 90 minutes. The time period of another satellite in an orbit at a distance of three times the radius of earth from the earth's surface will be
1 $90 \sqrt{8}$ minutes
2 360 minutes
3 720 minutes
4 270 minutes
Explanation:
C Given, $\mathrm{R}_{1}=\mathrm{R}_{\mathrm{e}}$ $\mathrm{R}_{2}=\mathrm{R}_{\mathrm{e}}+3 \mathrm{R}_{\mathrm{e}}=4 \mathrm{R}_{\mathrm{e}}$ $\mathrm{T}_{1}=90 \text { min. }$ From, Kepler's third law of planetary motion, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ So, $\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{2}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3}$ $\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{2}=\left(\frac{\mathrm{R}_{1}}{4 \mathrm{R}_{\mathrm{e}}}\right)^{3}$ $\frac{(90)^{2}}{\left(\mathrm{~T}_{2}\right)^{2}}=\frac{1}{64}$ $\mathrm{~T}_{2}=\sqrt{(90)^{2} \times 64}$ $\mathrm{~T}_{2}=90 \times 8$ $\mathrm{~T}_{2}=720$
AP EAMCET-28.04.2017
Gravitation
138652
A small satellite is in elliptical orbit around the earth as shown. $L$ denotes the magnitude of its angular momentum and $K$ denotes its kinetic energy. 1 and 2 denote two positions of the satellite, then
B By kepler's law, when a satellite is moving around the earth on elliptical path, then its angular momentum remains constant. $\mathrm{L}=\operatorname{mvr} \text { (angular momentum) }$ $\mathrm{K}=\frac{1}{2} \mathrm{mv}^2 \text { (Kinetic energy) }$ $\because \quad \mathrm{L}_1=\mathrm{L}_2$ $\mathrm{~m}_1 \mathrm{v}_1 \mathrm{r}_1=\mathrm{m}_2 \mathrm{v}_2 \mathrm{r}_2$ $\text { But, } \quad \mathrm{m}_1=\mathrm{m}_2=\mathrm{m}$ $\therefore \quad \mathrm{v}_1 \mathrm{r}_1=\mathrm{v}_2 \mathrm{r}_2$ $\frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{\mathrm{v}_2}{\mathrm{v}_1}$ $\text { Here, } \quad \mathrm{r}_1>\mathrm{r}_2$ $\frac{\mathrm{r}_1}{\mathrm{r}_2}>1$ $\frac{\mathrm{v}_2}{\mathrm{v}_1}>1$ $\mathrm{v}_2>\mathrm{v}_1$ $\mathrm{v}_2^2>\mathrm{v}_1^2 \text { or } \frac{1}{2} \mathrm{mv}_2^2>\frac{1}{2} \mathrm{mv}_1^2$ $\mathrm{~K}_2>\mathrm{K}_1\left(\because \mathrm{K}=\frac{1}{2} \mathrm{mv}^2\right)$ From equation (i), \(\frac{\mathrm{v}_2}{\mathrm{v}_1}>1\) \(\mathrm{v}_2>\mathrm{v}_1\) \(\mathrm{v}_2^2>\mathrm{v}_1^2 \text { or } \frac{1}{2} \mathrm{mv}_2^2>\frac{1}{2} \mathrm{mv}_1^2\) \(\mathrm{~K}_2>\mathrm{K}_1\left(\because \mathrm{K}=\frac{1}{2} \mathrm{mv}^2\right)\)
AMU-2018
Gravitation
138653
A planet is moving in an elliptical orbit around the sun. If $T, U, E$ and $L$ stand for its kinetic energy, gravitational potential energy, total energy and magnitude of angular momentum about the centre of force, which of the following is correct?
1 $\mathrm{T}$ is conserved
2 $\mathrm{U}$ is always positive
3 $\mathrm{E}$ is always negative
4 $\mathrm{L}$ is conserved but direction of vector $\mathrm{L}$ changes continuously
Explanation:
C Angular momentum conserved with direction of $\overrightarrow{\mathrm{L}}$ always in one direction. $\overrightarrow{\mathrm{L}}=\mathrm{mvr}=\text { constant }$ $\text { K.E. }=\mathrm{T}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{U}=-\frac{\mathrm{GMm}}{\mathrm{r}}$ For the planet to move in any orbit, its total mechanical energy must be bound then $\mathrm{E}$ must be negative always. Kinetic energy (T):- The path of the planet is given to be elliptical, so the distance between the center and the planet will change. Due to this change, As a result, the kinetic energy will not be conserved. Potential energy (U): For any attractive field, the potential energy is negative. Since the work is done against the acceleration, hence there is always a negative sign potential energy.
138650
Assume a planet with orbiting radius $R$ and period of revolution $T$ around the sun experiences a gravitational force which follows inverse cube law instead of inverse square law. In this case, the period of revolution is proportional to
1 $\frac{1}{\mathrm{R}}$
2 $\mathrm{R}^{2}$
3 $\mathrm{R}^{3 / 2}$
4 $\mathrm{R}^{4}$
Explanation:
B If the law of gravitation becomes an inverse cube law instead of inverse square law, $\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{R}^{3}}$ For a planet of mass ' $\mathrm{m}$ ' revolving around the sun, we can write $\frac{\mathrm{GMm}}{\mathrm{R}^{3}}=\frac{\mathrm{mv}^{2}}{\mathrm{R}}$ $\mathrm{v}=\frac{\sqrt{\mathrm{GM}}}{\mathrm{R}}$ Time period of revolution of planet $\mathrm{T} =\frac{2 \pi R}{\mathrm{~V}}$ $\mathrm{~T} =\frac{2 \pi \mathrm{R}}{\frac{\sqrt{\mathrm{Gm}}}{\mathrm{R}}}$ $\mathrm{T} =\frac{2 \pi \mathrm{R}^{2}}{\sqrt{\mathrm{Gm}}}$ $\mathrm{T} \propto \mathrm{R}^{2}$
AP EAMCET-11.07.2022
Gravitation
138651
The period of revolution of an early satellite close to the surface of earth is 90 minutes. The time period of another satellite in an orbit at a distance of three times the radius of earth from the earth's surface will be
1 $90 \sqrt{8}$ minutes
2 360 minutes
3 720 minutes
4 270 minutes
Explanation:
C Given, $\mathrm{R}_{1}=\mathrm{R}_{\mathrm{e}}$ $\mathrm{R}_{2}=\mathrm{R}_{\mathrm{e}}+3 \mathrm{R}_{\mathrm{e}}=4 \mathrm{R}_{\mathrm{e}}$ $\mathrm{T}_{1}=90 \text { min. }$ From, Kepler's third law of planetary motion, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ So, $\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{2}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3}$ $\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{2}=\left(\frac{\mathrm{R}_{1}}{4 \mathrm{R}_{\mathrm{e}}}\right)^{3}$ $\frac{(90)^{2}}{\left(\mathrm{~T}_{2}\right)^{2}}=\frac{1}{64}$ $\mathrm{~T}_{2}=\sqrt{(90)^{2} \times 64}$ $\mathrm{~T}_{2}=90 \times 8$ $\mathrm{~T}_{2}=720$
AP EAMCET-28.04.2017
Gravitation
138652
A small satellite is in elliptical orbit around the earth as shown. $L$ denotes the magnitude of its angular momentum and $K$ denotes its kinetic energy. 1 and 2 denote two positions of the satellite, then
B By kepler's law, when a satellite is moving around the earth on elliptical path, then its angular momentum remains constant. $\mathrm{L}=\operatorname{mvr} \text { (angular momentum) }$ $\mathrm{K}=\frac{1}{2} \mathrm{mv}^2 \text { (Kinetic energy) }$ $\because \quad \mathrm{L}_1=\mathrm{L}_2$ $\mathrm{~m}_1 \mathrm{v}_1 \mathrm{r}_1=\mathrm{m}_2 \mathrm{v}_2 \mathrm{r}_2$ $\text { But, } \quad \mathrm{m}_1=\mathrm{m}_2=\mathrm{m}$ $\therefore \quad \mathrm{v}_1 \mathrm{r}_1=\mathrm{v}_2 \mathrm{r}_2$ $\frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{\mathrm{v}_2}{\mathrm{v}_1}$ $\text { Here, } \quad \mathrm{r}_1>\mathrm{r}_2$ $\frac{\mathrm{r}_1}{\mathrm{r}_2}>1$ $\frac{\mathrm{v}_2}{\mathrm{v}_1}>1$ $\mathrm{v}_2>\mathrm{v}_1$ $\mathrm{v}_2^2>\mathrm{v}_1^2 \text { or } \frac{1}{2} \mathrm{mv}_2^2>\frac{1}{2} \mathrm{mv}_1^2$ $\mathrm{~K}_2>\mathrm{K}_1\left(\because \mathrm{K}=\frac{1}{2} \mathrm{mv}^2\right)$ From equation (i), \(\frac{\mathrm{v}_2}{\mathrm{v}_1}>1\) \(\mathrm{v}_2>\mathrm{v}_1\) \(\mathrm{v}_2^2>\mathrm{v}_1^2 \text { or } \frac{1}{2} \mathrm{mv}_2^2>\frac{1}{2} \mathrm{mv}_1^2\) \(\mathrm{~K}_2>\mathrm{K}_1\left(\because \mathrm{K}=\frac{1}{2} \mathrm{mv}^2\right)\)
AMU-2018
Gravitation
138653
A planet is moving in an elliptical orbit around the sun. If $T, U, E$ and $L$ stand for its kinetic energy, gravitational potential energy, total energy and magnitude of angular momentum about the centre of force, which of the following is correct?
1 $\mathrm{T}$ is conserved
2 $\mathrm{U}$ is always positive
3 $\mathrm{E}$ is always negative
4 $\mathrm{L}$ is conserved but direction of vector $\mathrm{L}$ changes continuously
Explanation:
C Angular momentum conserved with direction of $\overrightarrow{\mathrm{L}}$ always in one direction. $\overrightarrow{\mathrm{L}}=\mathrm{mvr}=\text { constant }$ $\text { K.E. }=\mathrm{T}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{U}=-\frac{\mathrm{GMm}}{\mathrm{r}}$ For the planet to move in any orbit, its total mechanical energy must be bound then $\mathrm{E}$ must be negative always. Kinetic energy (T):- The path of the planet is given to be elliptical, so the distance between the center and the planet will change. Due to this change, As a result, the kinetic energy will not be conserved. Potential energy (U): For any attractive field, the potential energy is negative. Since the work is done against the acceleration, hence there is always a negative sign potential energy.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Gravitation
138650
Assume a planet with orbiting radius $R$ and period of revolution $T$ around the sun experiences a gravitational force which follows inverse cube law instead of inverse square law. In this case, the period of revolution is proportional to
1 $\frac{1}{\mathrm{R}}$
2 $\mathrm{R}^{2}$
3 $\mathrm{R}^{3 / 2}$
4 $\mathrm{R}^{4}$
Explanation:
B If the law of gravitation becomes an inverse cube law instead of inverse square law, $\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{R}^{3}}$ For a planet of mass ' $\mathrm{m}$ ' revolving around the sun, we can write $\frac{\mathrm{GMm}}{\mathrm{R}^{3}}=\frac{\mathrm{mv}^{2}}{\mathrm{R}}$ $\mathrm{v}=\frac{\sqrt{\mathrm{GM}}}{\mathrm{R}}$ Time period of revolution of planet $\mathrm{T} =\frac{2 \pi R}{\mathrm{~V}}$ $\mathrm{~T} =\frac{2 \pi \mathrm{R}}{\frac{\sqrt{\mathrm{Gm}}}{\mathrm{R}}}$ $\mathrm{T} =\frac{2 \pi \mathrm{R}^{2}}{\sqrt{\mathrm{Gm}}}$ $\mathrm{T} \propto \mathrm{R}^{2}$
AP EAMCET-11.07.2022
Gravitation
138651
The period of revolution of an early satellite close to the surface of earth is 90 minutes. The time period of another satellite in an orbit at a distance of three times the radius of earth from the earth's surface will be
1 $90 \sqrt{8}$ minutes
2 360 minutes
3 720 minutes
4 270 minutes
Explanation:
C Given, $\mathrm{R}_{1}=\mathrm{R}_{\mathrm{e}}$ $\mathrm{R}_{2}=\mathrm{R}_{\mathrm{e}}+3 \mathrm{R}_{\mathrm{e}}=4 \mathrm{R}_{\mathrm{e}}$ $\mathrm{T}_{1}=90 \text { min. }$ From, Kepler's third law of planetary motion, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ So, $\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{2}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3}$ $\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{2}=\left(\frac{\mathrm{R}_{1}}{4 \mathrm{R}_{\mathrm{e}}}\right)^{3}$ $\frac{(90)^{2}}{\left(\mathrm{~T}_{2}\right)^{2}}=\frac{1}{64}$ $\mathrm{~T}_{2}=\sqrt{(90)^{2} \times 64}$ $\mathrm{~T}_{2}=90 \times 8$ $\mathrm{~T}_{2}=720$
AP EAMCET-28.04.2017
Gravitation
138652
A small satellite is in elliptical orbit around the earth as shown. $L$ denotes the magnitude of its angular momentum and $K$ denotes its kinetic energy. 1 and 2 denote two positions of the satellite, then
B By kepler's law, when a satellite is moving around the earth on elliptical path, then its angular momentum remains constant. $\mathrm{L}=\operatorname{mvr} \text { (angular momentum) }$ $\mathrm{K}=\frac{1}{2} \mathrm{mv}^2 \text { (Kinetic energy) }$ $\because \quad \mathrm{L}_1=\mathrm{L}_2$ $\mathrm{~m}_1 \mathrm{v}_1 \mathrm{r}_1=\mathrm{m}_2 \mathrm{v}_2 \mathrm{r}_2$ $\text { But, } \quad \mathrm{m}_1=\mathrm{m}_2=\mathrm{m}$ $\therefore \quad \mathrm{v}_1 \mathrm{r}_1=\mathrm{v}_2 \mathrm{r}_2$ $\frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{\mathrm{v}_2}{\mathrm{v}_1}$ $\text { Here, } \quad \mathrm{r}_1>\mathrm{r}_2$ $\frac{\mathrm{r}_1}{\mathrm{r}_2}>1$ $\frac{\mathrm{v}_2}{\mathrm{v}_1}>1$ $\mathrm{v}_2>\mathrm{v}_1$ $\mathrm{v}_2^2>\mathrm{v}_1^2 \text { or } \frac{1}{2} \mathrm{mv}_2^2>\frac{1}{2} \mathrm{mv}_1^2$ $\mathrm{~K}_2>\mathrm{K}_1\left(\because \mathrm{K}=\frac{1}{2} \mathrm{mv}^2\right)$ From equation (i), \(\frac{\mathrm{v}_2}{\mathrm{v}_1}>1\) \(\mathrm{v}_2>\mathrm{v}_1\) \(\mathrm{v}_2^2>\mathrm{v}_1^2 \text { or } \frac{1}{2} \mathrm{mv}_2^2>\frac{1}{2} \mathrm{mv}_1^2\) \(\mathrm{~K}_2>\mathrm{K}_1\left(\because \mathrm{K}=\frac{1}{2} \mathrm{mv}^2\right)\)
AMU-2018
Gravitation
138653
A planet is moving in an elliptical orbit around the sun. If $T, U, E$ and $L$ stand for its kinetic energy, gravitational potential energy, total energy and magnitude of angular momentum about the centre of force, which of the following is correct?
1 $\mathrm{T}$ is conserved
2 $\mathrm{U}$ is always positive
3 $\mathrm{E}$ is always negative
4 $\mathrm{L}$ is conserved but direction of vector $\mathrm{L}$ changes continuously
Explanation:
C Angular momentum conserved with direction of $\overrightarrow{\mathrm{L}}$ always in one direction. $\overrightarrow{\mathrm{L}}=\mathrm{mvr}=\text { constant }$ $\text { K.E. }=\mathrm{T}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{U}=-\frac{\mathrm{GMm}}{\mathrm{r}}$ For the planet to move in any orbit, its total mechanical energy must be bound then $\mathrm{E}$ must be negative always. Kinetic energy (T):- The path of the planet is given to be elliptical, so the distance between the center and the planet will change. Due to this change, As a result, the kinetic energy will not be conserved. Potential energy (U): For any attractive field, the potential energy is negative. Since the work is done against the acceleration, hence there is always a negative sign potential energy.
