138625
Kepler's second law states that the radius vector to a planet from the sun sweeps out equal areas in equal intervals of time. This law is a consequence of the conservation of:
1 Time
2 Mass
3 Angular momentum
4 Linear momentum
Explanation:
C Kepler's second law is a consequence of the conservation of angular momentum. Areal velocity $\left(\frac{\Delta \mathrm{A}}{\Delta \mathrm{t}}\right)=\frac{\mathrm{L}}{2 \mathrm{~m}}$ Since, the radius vector of a planet from the sun sweeps out equal area in equal interval of time, $\frac{\Delta \mathrm{A}}{\Delta \mathrm{t}}=\mathrm{constant}$ $\therefore \quad \mathrm{L}=$ constant [angular momentum (L) of planet about the centre of sun]
TS EAMCET(Medical)-2017
Gravitation
138626
The period $T$ and radius $R$ of a circular orbit of a planet about the sun are related by
1 $T^{2} \propto R^{3}$
2 $\mathrm{T} \propto \mathrm{R}$
3 $\mathrm{T}^{2} \propto \sqrt{\mathrm{R}^{3}}$
4 $\mathrm{T}^{2} \propto \mathrm{R}^{4}$
Explanation:
A We Know that the Kepler's third law, gives the period of a circular orbit of radius $\mathrm{R}$ about earth. $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{GM}_{\mathrm{e}}}}$ Where, $\mathrm{M}_{\mathrm{e}}=\text { mass of the earth }$ On squaring both side- $\mathrm{T}^{2}=(2 \pi)^{2} \frac{\mathrm{R}^{3}}{\mathrm{GM}_{\mathrm{e}}}$ $\mathrm{T}^{2}=\left(\frac{4 \pi^{2}}{\mathrm{GM}_{\mathrm{e}}}\right) \mathrm{R}^{3}$ $\operatorname{As}\left(\frac{4 \pi^{2}}{\mathrm{GM}_{\mathrm{e}}}\right) \text { is a constant }$ $\therefore \quad \mathrm{T}^{2} \propto \mathrm{R}^{3}$
TS EAMCET 29.09.2020
Gravitation
138627
The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become.
1 10 hours
2 80 hours
3 40 hours
4 20 hours
Explanation:
C Given, The time period of satellite of earth $(T)=5$ hours, Radius of earth $=\mathrm{R}$ Radius of planet $=4 \mathrm{R}$ Time period of satellite of planet $\left(\mathrm{T}^{\prime}\right)=$ ? We know that, $\mathrm{T}^{2} \propto \mathrm{r}^{3} \quad \text { (Kepler's third law) }$ Now, $\left(\frac{\mathrm{T}^{\prime}}{\mathrm{T}}\right)^{2}=\left(\frac{\mathrm{R}^{\prime}}{\mathrm{R}}\right)^{3}$ $\mathrm{~T}^{\prime}=\left(\frac{\mathrm{R}^{\prime}}{\mathrm{R}}\right)^{3 / 2} \times \mathrm{T}=(4)^{3 / 2} \times 5=8 \times 5$ $\mathrm{~T}^{\prime}=40 \text { hours }$
SRMJEE - 2007
Gravitation
138628
If a planet revolves around the sun in a circular orbit of radius "a" with a period of revolution $T$, then ( $K$ being a positive constant)
1 $\mathrm{T}=\mathrm{Ka}^{2 / 3}$
2 $\mathrm{T}=\mathrm{Ka}^{3 / 2}$
3 $\mathrm{T}=\mathrm{Ka}^{2}$
4 $\mathrm{T}=\mathrm{Ka}^{3}$
Explanation:
B $M=$ mass of sun Gravitational force $=$ centrifugal force $\quad \frac{G M m}{a^{2}}=m \omega^{2} a$ $\omega^{2}=\frac{G M}{a^{3}}$ $\omega=\frac{\sqrt{\mathrm{GM}}}{(\mathrm{a})^{3 / 2}}$ $\because \quad \mathrm{T}=\frac{2 \pi}{\omega}$ $\mathrm{T}=\frac{2 \pi}{\sqrt{\mathrm{GM}}} \times \mathrm{a}^{3 / 2}$ $\mathrm{~T}=\mathrm{K} \times \mathrm{a}^{3 / 2}$ $\mathrm{~T}=\mathrm{Ka}^{3 / 2}$
138625
Kepler's second law states that the radius vector to a planet from the sun sweeps out equal areas in equal intervals of time. This law is a consequence of the conservation of:
1 Time
2 Mass
3 Angular momentum
4 Linear momentum
Explanation:
C Kepler's second law is a consequence of the conservation of angular momentum. Areal velocity $\left(\frac{\Delta \mathrm{A}}{\Delta \mathrm{t}}\right)=\frac{\mathrm{L}}{2 \mathrm{~m}}$ Since, the radius vector of a planet from the sun sweeps out equal area in equal interval of time, $\frac{\Delta \mathrm{A}}{\Delta \mathrm{t}}=\mathrm{constant}$ $\therefore \quad \mathrm{L}=$ constant [angular momentum (L) of planet about the centre of sun]
TS EAMCET(Medical)-2017
Gravitation
138626
The period $T$ and radius $R$ of a circular orbit of a planet about the sun are related by
1 $T^{2} \propto R^{3}$
2 $\mathrm{T} \propto \mathrm{R}$
3 $\mathrm{T}^{2} \propto \sqrt{\mathrm{R}^{3}}$
4 $\mathrm{T}^{2} \propto \mathrm{R}^{4}$
Explanation:
A We Know that the Kepler's third law, gives the period of a circular orbit of radius $\mathrm{R}$ about earth. $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{GM}_{\mathrm{e}}}}$ Where, $\mathrm{M}_{\mathrm{e}}=\text { mass of the earth }$ On squaring both side- $\mathrm{T}^{2}=(2 \pi)^{2} \frac{\mathrm{R}^{3}}{\mathrm{GM}_{\mathrm{e}}}$ $\mathrm{T}^{2}=\left(\frac{4 \pi^{2}}{\mathrm{GM}_{\mathrm{e}}}\right) \mathrm{R}^{3}$ $\operatorname{As}\left(\frac{4 \pi^{2}}{\mathrm{GM}_{\mathrm{e}}}\right) \text { is a constant }$ $\therefore \quad \mathrm{T}^{2} \propto \mathrm{R}^{3}$
TS EAMCET 29.09.2020
Gravitation
138627
The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become.
1 10 hours
2 80 hours
3 40 hours
4 20 hours
Explanation:
C Given, The time period of satellite of earth $(T)=5$ hours, Radius of earth $=\mathrm{R}$ Radius of planet $=4 \mathrm{R}$ Time period of satellite of planet $\left(\mathrm{T}^{\prime}\right)=$ ? We know that, $\mathrm{T}^{2} \propto \mathrm{r}^{3} \quad \text { (Kepler's third law) }$ Now, $\left(\frac{\mathrm{T}^{\prime}}{\mathrm{T}}\right)^{2}=\left(\frac{\mathrm{R}^{\prime}}{\mathrm{R}}\right)^{3}$ $\mathrm{~T}^{\prime}=\left(\frac{\mathrm{R}^{\prime}}{\mathrm{R}}\right)^{3 / 2} \times \mathrm{T}=(4)^{3 / 2} \times 5=8 \times 5$ $\mathrm{~T}^{\prime}=40 \text { hours }$
SRMJEE - 2007
Gravitation
138628
If a planet revolves around the sun in a circular orbit of radius "a" with a period of revolution $T$, then ( $K$ being a positive constant)
1 $\mathrm{T}=\mathrm{Ka}^{2 / 3}$
2 $\mathrm{T}=\mathrm{Ka}^{3 / 2}$
3 $\mathrm{T}=\mathrm{Ka}^{2}$
4 $\mathrm{T}=\mathrm{Ka}^{3}$
Explanation:
B $M=$ mass of sun Gravitational force $=$ centrifugal force $\quad \frac{G M m}{a^{2}}=m \omega^{2} a$ $\omega^{2}=\frac{G M}{a^{3}}$ $\omega=\frac{\sqrt{\mathrm{GM}}}{(\mathrm{a})^{3 / 2}}$ $\because \quad \mathrm{T}=\frac{2 \pi}{\omega}$ $\mathrm{T}=\frac{2 \pi}{\sqrt{\mathrm{GM}}} \times \mathrm{a}^{3 / 2}$ $\mathrm{~T}=\mathrm{K} \times \mathrm{a}^{3 / 2}$ $\mathrm{~T}=\mathrm{Ka}^{3 / 2}$
138625
Kepler's second law states that the radius vector to a planet from the sun sweeps out equal areas in equal intervals of time. This law is a consequence of the conservation of:
1 Time
2 Mass
3 Angular momentum
4 Linear momentum
Explanation:
C Kepler's second law is a consequence of the conservation of angular momentum. Areal velocity $\left(\frac{\Delta \mathrm{A}}{\Delta \mathrm{t}}\right)=\frac{\mathrm{L}}{2 \mathrm{~m}}$ Since, the radius vector of a planet from the sun sweeps out equal area in equal interval of time, $\frac{\Delta \mathrm{A}}{\Delta \mathrm{t}}=\mathrm{constant}$ $\therefore \quad \mathrm{L}=$ constant [angular momentum (L) of planet about the centre of sun]
TS EAMCET(Medical)-2017
Gravitation
138626
The period $T$ and radius $R$ of a circular orbit of a planet about the sun are related by
1 $T^{2} \propto R^{3}$
2 $\mathrm{T} \propto \mathrm{R}$
3 $\mathrm{T}^{2} \propto \sqrt{\mathrm{R}^{3}}$
4 $\mathrm{T}^{2} \propto \mathrm{R}^{4}$
Explanation:
A We Know that the Kepler's third law, gives the period of a circular orbit of radius $\mathrm{R}$ about earth. $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{GM}_{\mathrm{e}}}}$ Where, $\mathrm{M}_{\mathrm{e}}=\text { mass of the earth }$ On squaring both side- $\mathrm{T}^{2}=(2 \pi)^{2} \frac{\mathrm{R}^{3}}{\mathrm{GM}_{\mathrm{e}}}$ $\mathrm{T}^{2}=\left(\frac{4 \pi^{2}}{\mathrm{GM}_{\mathrm{e}}}\right) \mathrm{R}^{3}$ $\operatorname{As}\left(\frac{4 \pi^{2}}{\mathrm{GM}_{\mathrm{e}}}\right) \text { is a constant }$ $\therefore \quad \mathrm{T}^{2} \propto \mathrm{R}^{3}$
TS EAMCET 29.09.2020
Gravitation
138627
The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become.
1 10 hours
2 80 hours
3 40 hours
4 20 hours
Explanation:
C Given, The time period of satellite of earth $(T)=5$ hours, Radius of earth $=\mathrm{R}$ Radius of planet $=4 \mathrm{R}$ Time period of satellite of planet $\left(\mathrm{T}^{\prime}\right)=$ ? We know that, $\mathrm{T}^{2} \propto \mathrm{r}^{3} \quad \text { (Kepler's third law) }$ Now, $\left(\frac{\mathrm{T}^{\prime}}{\mathrm{T}}\right)^{2}=\left(\frac{\mathrm{R}^{\prime}}{\mathrm{R}}\right)^{3}$ $\mathrm{~T}^{\prime}=\left(\frac{\mathrm{R}^{\prime}}{\mathrm{R}}\right)^{3 / 2} \times \mathrm{T}=(4)^{3 / 2} \times 5=8 \times 5$ $\mathrm{~T}^{\prime}=40 \text { hours }$
SRMJEE - 2007
Gravitation
138628
If a planet revolves around the sun in a circular orbit of radius "a" with a period of revolution $T$, then ( $K$ being a positive constant)
1 $\mathrm{T}=\mathrm{Ka}^{2 / 3}$
2 $\mathrm{T}=\mathrm{Ka}^{3 / 2}$
3 $\mathrm{T}=\mathrm{Ka}^{2}$
4 $\mathrm{T}=\mathrm{Ka}^{3}$
Explanation:
B $M=$ mass of sun Gravitational force $=$ centrifugal force $\quad \frac{G M m}{a^{2}}=m \omega^{2} a$ $\omega^{2}=\frac{G M}{a^{3}}$ $\omega=\frac{\sqrt{\mathrm{GM}}}{(\mathrm{a})^{3 / 2}}$ $\because \quad \mathrm{T}=\frac{2 \pi}{\omega}$ $\mathrm{T}=\frac{2 \pi}{\sqrt{\mathrm{GM}}} \times \mathrm{a}^{3 / 2}$ $\mathrm{~T}=\mathrm{K} \times \mathrm{a}^{3 / 2}$ $\mathrm{~T}=\mathrm{Ka}^{3 / 2}$
138625
Kepler's second law states that the radius vector to a planet from the sun sweeps out equal areas in equal intervals of time. This law is a consequence of the conservation of:
1 Time
2 Mass
3 Angular momentum
4 Linear momentum
Explanation:
C Kepler's second law is a consequence of the conservation of angular momentum. Areal velocity $\left(\frac{\Delta \mathrm{A}}{\Delta \mathrm{t}}\right)=\frac{\mathrm{L}}{2 \mathrm{~m}}$ Since, the radius vector of a planet from the sun sweeps out equal area in equal interval of time, $\frac{\Delta \mathrm{A}}{\Delta \mathrm{t}}=\mathrm{constant}$ $\therefore \quad \mathrm{L}=$ constant [angular momentum (L) of planet about the centre of sun]
TS EAMCET(Medical)-2017
Gravitation
138626
The period $T$ and radius $R$ of a circular orbit of a planet about the sun are related by
1 $T^{2} \propto R^{3}$
2 $\mathrm{T} \propto \mathrm{R}$
3 $\mathrm{T}^{2} \propto \sqrt{\mathrm{R}^{3}}$
4 $\mathrm{T}^{2} \propto \mathrm{R}^{4}$
Explanation:
A We Know that the Kepler's third law, gives the period of a circular orbit of radius $\mathrm{R}$ about earth. $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{GM}_{\mathrm{e}}}}$ Where, $\mathrm{M}_{\mathrm{e}}=\text { mass of the earth }$ On squaring both side- $\mathrm{T}^{2}=(2 \pi)^{2} \frac{\mathrm{R}^{3}}{\mathrm{GM}_{\mathrm{e}}}$ $\mathrm{T}^{2}=\left(\frac{4 \pi^{2}}{\mathrm{GM}_{\mathrm{e}}}\right) \mathrm{R}^{3}$ $\operatorname{As}\left(\frac{4 \pi^{2}}{\mathrm{GM}_{\mathrm{e}}}\right) \text { is a constant }$ $\therefore \quad \mathrm{T}^{2} \propto \mathrm{R}^{3}$
TS EAMCET 29.09.2020
Gravitation
138627
The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become.
1 10 hours
2 80 hours
3 40 hours
4 20 hours
Explanation:
C Given, The time period of satellite of earth $(T)=5$ hours, Radius of earth $=\mathrm{R}$ Radius of planet $=4 \mathrm{R}$ Time period of satellite of planet $\left(\mathrm{T}^{\prime}\right)=$ ? We know that, $\mathrm{T}^{2} \propto \mathrm{r}^{3} \quad \text { (Kepler's third law) }$ Now, $\left(\frac{\mathrm{T}^{\prime}}{\mathrm{T}}\right)^{2}=\left(\frac{\mathrm{R}^{\prime}}{\mathrm{R}}\right)^{3}$ $\mathrm{~T}^{\prime}=\left(\frac{\mathrm{R}^{\prime}}{\mathrm{R}}\right)^{3 / 2} \times \mathrm{T}=(4)^{3 / 2} \times 5=8 \times 5$ $\mathrm{~T}^{\prime}=40 \text { hours }$
SRMJEE - 2007
Gravitation
138628
If a planet revolves around the sun in a circular orbit of radius "a" with a period of revolution $T$, then ( $K$ being a positive constant)
1 $\mathrm{T}=\mathrm{Ka}^{2 / 3}$
2 $\mathrm{T}=\mathrm{Ka}^{3 / 2}$
3 $\mathrm{T}=\mathrm{Ka}^{2}$
4 $\mathrm{T}=\mathrm{Ka}^{3}$
Explanation:
B $M=$ mass of sun Gravitational force $=$ centrifugal force $\quad \frac{G M m}{a^{2}}=m \omega^{2} a$ $\omega^{2}=\frac{G M}{a^{3}}$ $\omega=\frac{\sqrt{\mathrm{GM}}}{(\mathrm{a})^{3 / 2}}$ $\because \quad \mathrm{T}=\frac{2 \pi}{\omega}$ $\mathrm{T}=\frac{2 \pi}{\sqrt{\mathrm{GM}}} \times \mathrm{a}^{3 / 2}$ $\mathrm{~T}=\mathrm{K} \times \mathrm{a}^{3 / 2}$ $\mathrm{~T}=\mathrm{Ka}^{3 / 2}$
