138620
Time period of an artificial satellite close to the spherical planet of radius ' $R$ ' is $T$. The period of revolution for the same satellite close to the surface of another planet of radius '3R' is (Density of both the planets is same.)
1 $\sqrt{3} \mathrm{~T}$
2 $\mathrm{T}$
3 $3 \mathrm{~T}$
4 $9 \mathrm{~T}$
Explanation:
B We know that, Time period of satellite $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{GM}}}$ Density of both planets is same $=\rho$ $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{G} \frac{4}{3} \pi \mathrm{R}^{3} \rho}}$ $\mathrm{T}=2 \pi \sqrt{\frac{3}{4 \mathrm{G} \rho \pi}}$ $\mathrm{T} \propto \frac{1}{\sqrt{\rho}}$ The time period of artificial satellite depend only density of the planet. As, density of both planet are same so time period of both satellite will be same.
MHT-CET 2019
Gravitation
138621
Kepler's second law states that the straight line joining the planet to the sun sweeps out equal areas in equal times. This statement is equivalent to saying that.
1 total acceleration is zero
2 tangential acceleration is zero
3 longitudinal acceleration is zero
4 radial acceleration is zero
Explanation:
B Kepler's second law states that the straight line joining the planet to the sun sweeps out equal areas in equal times. Means that areal velocity of the planet is constant. $\therefore \quad \frac{\Delta \overrightarrow{\mathrm{A}}}{\Delta \mathrm{t}}=\frac{\overrightarrow{\mathrm{L}}}{2 \mathrm{M}}=\mathrm{a} \text { constant vector }$ $\tau =\frac{\Delta \mathrm{L}}{\Delta \mathrm{t}}=0$ $0 =\frac{\mathrm{I} \Delta \omega}{\Delta \mathrm{t}} \quad(\mathrm{L}=\mathrm{I} \omega)$ $0 =\mathrm{I} \alpha$ $\alpha =0 \quad\left(\frac{\Delta \omega}{\Delta \mathrm{t}}=\alpha\right)$ $\therefore$ Tangential acceleration $\mathrm{a}_{\mathrm{t}}=\mathrm{r} \alpha=0$ The given statement is equivalent to saying that tangential acceleration is zero.
MHT-CET 2006
Gravitation
138623
If distance between earth and sun become four times, then time period becomes.
1 4 times
2 8 times
3 $1 / 4$ times
4 $1 / 8$ times
Explanation:
B We know that, $\mathrm{T}^{2} \propto \mathrm{R}^{3} \text { (Keplar's third law) }$ $\mathrm{T}^{2}=\mathrm{KR}^{3} \ldots . \text { (i) }$ Where, $\mathrm{T}$ is time taken by the planet to revolve around the sun. $\mathrm{R}$ is semi-major axis of ellipse. $\mathrm{K}$ is constant of proportionality. According to question, $\mathrm{R}^{\prime}=4 \mathrm{R}$ $\therefore \mathrm{T}^{\prime 2}=\mathrm{KR}^{\prime 3}=\mathrm{K}(4 \mathrm{R})^{3} \quad \ldots . \text { (ii) }$ Now, on dividing equation (i) and (ii), we get - $\frac{\mathrm{T}^{2}}{\mathrm{~T}^{\prime 2}}=\frac{\mathrm{KR}^{3}}{\mathrm{~K}(4 \mathrm{R})^{3}}$ $\frac{\mathrm{T}^{2}}{\mathrm{~T}^{\prime 2}}=\frac{\mathrm{KR}^{3}}{64 \mathrm{KR}^{3}}$ $\left(\frac{\mathrm{T}}{\mathrm{T}^{\prime}}\right)^{2}=\frac{1}{64}$ $\frac{\mathrm{T}}{\mathrm{T}^{\prime}}=\sqrt{\frac{1}{64}}$ $\frac{\mathrm{T}}{\mathrm{T}^{\prime}}=\frac{1}{8}$ $\mathrm{~T}^{\prime}=8 \mathrm{~T}$ So, time period becomes 8 times of pervious value.
UP CPMT-2007
Gravitation
138624
In the adjoining figure a planet $m$ revolves in elliptical orbit about the sun $S$. The shaded area SCD is twice that of shaded area SAB. If $t_{1}$ is the time for the planet to move from $C$ to $D$ and $t_{2}$ is the time to move from $A$ to $B$, then
1 $t_{1}>t_{2}$
2 $\mathrm{t}_{1}+\mathrm{t}_{2}$
3 $\mathrm{t}_{1}=4 \mathrm{t}_{2}$
4 $\mathrm{t}_{1}=2 \mathrm{t}_{2}$
Explanation:
D According to Kepler's law of area (Second Law), the areal velocity of a Planet around the sun always remains constant. According to the question- SCD: Area $\left(A_{1}\right)$ swept by planet in time interval $t_{1}$. SAB : Area $\left(\mathrm{A}_{2}\right)$ swept by planet in time interval $\mathrm{t}_{2}$. Given, $\mathrm{A}_{1}=2 \mathrm{~A}_{2}$ As areal velocity is constant. So, $\frac{A_{1}}{t_{1}} =\frac{A_{2}}{t_{2}}$ $t_{1} =t_{2} \cdot \frac{A_{1}}{A_{2}}$ $t_{1} =t_{2} \cdot \frac{2 A_{2}}{A_{2}}$ $t_{1} =2 t_{2}$
138620
Time period of an artificial satellite close to the spherical planet of radius ' $R$ ' is $T$. The period of revolution for the same satellite close to the surface of another planet of radius '3R' is (Density of both the planets is same.)
1 $\sqrt{3} \mathrm{~T}$
2 $\mathrm{T}$
3 $3 \mathrm{~T}$
4 $9 \mathrm{~T}$
Explanation:
B We know that, Time period of satellite $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{GM}}}$ Density of both planets is same $=\rho$ $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{G} \frac{4}{3} \pi \mathrm{R}^{3} \rho}}$ $\mathrm{T}=2 \pi \sqrt{\frac{3}{4 \mathrm{G} \rho \pi}}$ $\mathrm{T} \propto \frac{1}{\sqrt{\rho}}$ The time period of artificial satellite depend only density of the planet. As, density of both planet are same so time period of both satellite will be same.
MHT-CET 2019
Gravitation
138621
Kepler's second law states that the straight line joining the planet to the sun sweeps out equal areas in equal times. This statement is equivalent to saying that.
1 total acceleration is zero
2 tangential acceleration is zero
3 longitudinal acceleration is zero
4 radial acceleration is zero
Explanation:
B Kepler's second law states that the straight line joining the planet to the sun sweeps out equal areas in equal times. Means that areal velocity of the planet is constant. $\therefore \quad \frac{\Delta \overrightarrow{\mathrm{A}}}{\Delta \mathrm{t}}=\frac{\overrightarrow{\mathrm{L}}}{2 \mathrm{M}}=\mathrm{a} \text { constant vector }$ $\tau =\frac{\Delta \mathrm{L}}{\Delta \mathrm{t}}=0$ $0 =\frac{\mathrm{I} \Delta \omega}{\Delta \mathrm{t}} \quad(\mathrm{L}=\mathrm{I} \omega)$ $0 =\mathrm{I} \alpha$ $\alpha =0 \quad\left(\frac{\Delta \omega}{\Delta \mathrm{t}}=\alpha\right)$ $\therefore$ Tangential acceleration $\mathrm{a}_{\mathrm{t}}=\mathrm{r} \alpha=0$ The given statement is equivalent to saying that tangential acceleration is zero.
MHT-CET 2006
Gravitation
138623
If distance between earth and sun become four times, then time period becomes.
1 4 times
2 8 times
3 $1 / 4$ times
4 $1 / 8$ times
Explanation:
B We know that, $\mathrm{T}^{2} \propto \mathrm{R}^{3} \text { (Keplar's third law) }$ $\mathrm{T}^{2}=\mathrm{KR}^{3} \ldots . \text { (i) }$ Where, $\mathrm{T}$ is time taken by the planet to revolve around the sun. $\mathrm{R}$ is semi-major axis of ellipse. $\mathrm{K}$ is constant of proportionality. According to question, $\mathrm{R}^{\prime}=4 \mathrm{R}$ $\therefore \mathrm{T}^{\prime 2}=\mathrm{KR}^{\prime 3}=\mathrm{K}(4 \mathrm{R})^{3} \quad \ldots . \text { (ii) }$ Now, on dividing equation (i) and (ii), we get - $\frac{\mathrm{T}^{2}}{\mathrm{~T}^{\prime 2}}=\frac{\mathrm{KR}^{3}}{\mathrm{~K}(4 \mathrm{R})^{3}}$ $\frac{\mathrm{T}^{2}}{\mathrm{~T}^{\prime 2}}=\frac{\mathrm{KR}^{3}}{64 \mathrm{KR}^{3}}$ $\left(\frac{\mathrm{T}}{\mathrm{T}^{\prime}}\right)^{2}=\frac{1}{64}$ $\frac{\mathrm{T}}{\mathrm{T}^{\prime}}=\sqrt{\frac{1}{64}}$ $\frac{\mathrm{T}}{\mathrm{T}^{\prime}}=\frac{1}{8}$ $\mathrm{~T}^{\prime}=8 \mathrm{~T}$ So, time period becomes 8 times of pervious value.
UP CPMT-2007
Gravitation
138624
In the adjoining figure a planet $m$ revolves in elliptical orbit about the sun $S$. The shaded area SCD is twice that of shaded area SAB. If $t_{1}$ is the time for the planet to move from $C$ to $D$ and $t_{2}$ is the time to move from $A$ to $B$, then
1 $t_{1}>t_{2}$
2 $\mathrm{t}_{1}+\mathrm{t}_{2}$
3 $\mathrm{t}_{1}=4 \mathrm{t}_{2}$
4 $\mathrm{t}_{1}=2 \mathrm{t}_{2}$
Explanation:
D According to Kepler's law of area (Second Law), the areal velocity of a Planet around the sun always remains constant. According to the question- SCD: Area $\left(A_{1}\right)$ swept by planet in time interval $t_{1}$. SAB : Area $\left(\mathrm{A}_{2}\right)$ swept by planet in time interval $\mathrm{t}_{2}$. Given, $\mathrm{A}_{1}=2 \mathrm{~A}_{2}$ As areal velocity is constant. So, $\frac{A_{1}}{t_{1}} =\frac{A_{2}}{t_{2}}$ $t_{1} =t_{2} \cdot \frac{A_{1}}{A_{2}}$ $t_{1} =t_{2} \cdot \frac{2 A_{2}}{A_{2}}$ $t_{1} =2 t_{2}$
138620
Time period of an artificial satellite close to the spherical planet of radius ' $R$ ' is $T$. The period of revolution for the same satellite close to the surface of another planet of radius '3R' is (Density of both the planets is same.)
1 $\sqrt{3} \mathrm{~T}$
2 $\mathrm{T}$
3 $3 \mathrm{~T}$
4 $9 \mathrm{~T}$
Explanation:
B We know that, Time period of satellite $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{GM}}}$ Density of both planets is same $=\rho$ $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{G} \frac{4}{3} \pi \mathrm{R}^{3} \rho}}$ $\mathrm{T}=2 \pi \sqrt{\frac{3}{4 \mathrm{G} \rho \pi}}$ $\mathrm{T} \propto \frac{1}{\sqrt{\rho}}$ The time period of artificial satellite depend only density of the planet. As, density of both planet are same so time period of both satellite will be same.
MHT-CET 2019
Gravitation
138621
Kepler's second law states that the straight line joining the planet to the sun sweeps out equal areas in equal times. This statement is equivalent to saying that.
1 total acceleration is zero
2 tangential acceleration is zero
3 longitudinal acceleration is zero
4 radial acceleration is zero
Explanation:
B Kepler's second law states that the straight line joining the planet to the sun sweeps out equal areas in equal times. Means that areal velocity of the planet is constant. $\therefore \quad \frac{\Delta \overrightarrow{\mathrm{A}}}{\Delta \mathrm{t}}=\frac{\overrightarrow{\mathrm{L}}}{2 \mathrm{M}}=\mathrm{a} \text { constant vector }$ $\tau =\frac{\Delta \mathrm{L}}{\Delta \mathrm{t}}=0$ $0 =\frac{\mathrm{I} \Delta \omega}{\Delta \mathrm{t}} \quad(\mathrm{L}=\mathrm{I} \omega)$ $0 =\mathrm{I} \alpha$ $\alpha =0 \quad\left(\frac{\Delta \omega}{\Delta \mathrm{t}}=\alpha\right)$ $\therefore$ Tangential acceleration $\mathrm{a}_{\mathrm{t}}=\mathrm{r} \alpha=0$ The given statement is equivalent to saying that tangential acceleration is zero.
MHT-CET 2006
Gravitation
138623
If distance between earth and sun become four times, then time period becomes.
1 4 times
2 8 times
3 $1 / 4$ times
4 $1 / 8$ times
Explanation:
B We know that, $\mathrm{T}^{2} \propto \mathrm{R}^{3} \text { (Keplar's third law) }$ $\mathrm{T}^{2}=\mathrm{KR}^{3} \ldots . \text { (i) }$ Where, $\mathrm{T}$ is time taken by the planet to revolve around the sun. $\mathrm{R}$ is semi-major axis of ellipse. $\mathrm{K}$ is constant of proportionality. According to question, $\mathrm{R}^{\prime}=4 \mathrm{R}$ $\therefore \mathrm{T}^{\prime 2}=\mathrm{KR}^{\prime 3}=\mathrm{K}(4 \mathrm{R})^{3} \quad \ldots . \text { (ii) }$ Now, on dividing equation (i) and (ii), we get - $\frac{\mathrm{T}^{2}}{\mathrm{~T}^{\prime 2}}=\frac{\mathrm{KR}^{3}}{\mathrm{~K}(4 \mathrm{R})^{3}}$ $\frac{\mathrm{T}^{2}}{\mathrm{~T}^{\prime 2}}=\frac{\mathrm{KR}^{3}}{64 \mathrm{KR}^{3}}$ $\left(\frac{\mathrm{T}}{\mathrm{T}^{\prime}}\right)^{2}=\frac{1}{64}$ $\frac{\mathrm{T}}{\mathrm{T}^{\prime}}=\sqrt{\frac{1}{64}}$ $\frac{\mathrm{T}}{\mathrm{T}^{\prime}}=\frac{1}{8}$ $\mathrm{~T}^{\prime}=8 \mathrm{~T}$ So, time period becomes 8 times of pervious value.
UP CPMT-2007
Gravitation
138624
In the adjoining figure a planet $m$ revolves in elliptical orbit about the sun $S$. The shaded area SCD is twice that of shaded area SAB. If $t_{1}$ is the time for the planet to move from $C$ to $D$ and $t_{2}$ is the time to move from $A$ to $B$, then
1 $t_{1}>t_{2}$
2 $\mathrm{t}_{1}+\mathrm{t}_{2}$
3 $\mathrm{t}_{1}=4 \mathrm{t}_{2}$
4 $\mathrm{t}_{1}=2 \mathrm{t}_{2}$
Explanation:
D According to Kepler's law of area (Second Law), the areal velocity of a Planet around the sun always remains constant. According to the question- SCD: Area $\left(A_{1}\right)$ swept by planet in time interval $t_{1}$. SAB : Area $\left(\mathrm{A}_{2}\right)$ swept by planet in time interval $\mathrm{t}_{2}$. Given, $\mathrm{A}_{1}=2 \mathrm{~A}_{2}$ As areal velocity is constant. So, $\frac{A_{1}}{t_{1}} =\frac{A_{2}}{t_{2}}$ $t_{1} =t_{2} \cdot \frac{A_{1}}{A_{2}}$ $t_{1} =t_{2} \cdot \frac{2 A_{2}}{A_{2}}$ $t_{1} =2 t_{2}$
138620
Time period of an artificial satellite close to the spherical planet of radius ' $R$ ' is $T$. The period of revolution for the same satellite close to the surface of another planet of radius '3R' is (Density of both the planets is same.)
1 $\sqrt{3} \mathrm{~T}$
2 $\mathrm{T}$
3 $3 \mathrm{~T}$
4 $9 \mathrm{~T}$
Explanation:
B We know that, Time period of satellite $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{GM}}}$ Density of both planets is same $=\rho$ $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{G} \frac{4}{3} \pi \mathrm{R}^{3} \rho}}$ $\mathrm{T}=2 \pi \sqrt{\frac{3}{4 \mathrm{G} \rho \pi}}$ $\mathrm{T} \propto \frac{1}{\sqrt{\rho}}$ The time period of artificial satellite depend only density of the planet. As, density of both planet are same so time period of both satellite will be same.
MHT-CET 2019
Gravitation
138621
Kepler's second law states that the straight line joining the planet to the sun sweeps out equal areas in equal times. This statement is equivalent to saying that.
1 total acceleration is zero
2 tangential acceleration is zero
3 longitudinal acceleration is zero
4 radial acceleration is zero
Explanation:
B Kepler's second law states that the straight line joining the planet to the sun sweeps out equal areas in equal times. Means that areal velocity of the planet is constant. $\therefore \quad \frac{\Delta \overrightarrow{\mathrm{A}}}{\Delta \mathrm{t}}=\frac{\overrightarrow{\mathrm{L}}}{2 \mathrm{M}}=\mathrm{a} \text { constant vector }$ $\tau =\frac{\Delta \mathrm{L}}{\Delta \mathrm{t}}=0$ $0 =\frac{\mathrm{I} \Delta \omega}{\Delta \mathrm{t}} \quad(\mathrm{L}=\mathrm{I} \omega)$ $0 =\mathrm{I} \alpha$ $\alpha =0 \quad\left(\frac{\Delta \omega}{\Delta \mathrm{t}}=\alpha\right)$ $\therefore$ Tangential acceleration $\mathrm{a}_{\mathrm{t}}=\mathrm{r} \alpha=0$ The given statement is equivalent to saying that tangential acceleration is zero.
MHT-CET 2006
Gravitation
138623
If distance between earth and sun become four times, then time period becomes.
1 4 times
2 8 times
3 $1 / 4$ times
4 $1 / 8$ times
Explanation:
B We know that, $\mathrm{T}^{2} \propto \mathrm{R}^{3} \text { (Keplar's third law) }$ $\mathrm{T}^{2}=\mathrm{KR}^{3} \ldots . \text { (i) }$ Where, $\mathrm{T}$ is time taken by the planet to revolve around the sun. $\mathrm{R}$ is semi-major axis of ellipse. $\mathrm{K}$ is constant of proportionality. According to question, $\mathrm{R}^{\prime}=4 \mathrm{R}$ $\therefore \mathrm{T}^{\prime 2}=\mathrm{KR}^{\prime 3}=\mathrm{K}(4 \mathrm{R})^{3} \quad \ldots . \text { (ii) }$ Now, on dividing equation (i) and (ii), we get - $\frac{\mathrm{T}^{2}}{\mathrm{~T}^{\prime 2}}=\frac{\mathrm{KR}^{3}}{\mathrm{~K}(4 \mathrm{R})^{3}}$ $\frac{\mathrm{T}^{2}}{\mathrm{~T}^{\prime 2}}=\frac{\mathrm{KR}^{3}}{64 \mathrm{KR}^{3}}$ $\left(\frac{\mathrm{T}}{\mathrm{T}^{\prime}}\right)^{2}=\frac{1}{64}$ $\frac{\mathrm{T}}{\mathrm{T}^{\prime}}=\sqrt{\frac{1}{64}}$ $\frac{\mathrm{T}}{\mathrm{T}^{\prime}}=\frac{1}{8}$ $\mathrm{~T}^{\prime}=8 \mathrm{~T}$ So, time period becomes 8 times of pervious value.
UP CPMT-2007
Gravitation
138624
In the adjoining figure a planet $m$ revolves in elliptical orbit about the sun $S$. The shaded area SCD is twice that of shaded area SAB. If $t_{1}$ is the time for the planet to move from $C$ to $D$ and $t_{2}$ is the time to move from $A$ to $B$, then
1 $t_{1}>t_{2}$
2 $\mathrm{t}_{1}+\mathrm{t}_{2}$
3 $\mathrm{t}_{1}=4 \mathrm{t}_{2}$
4 $\mathrm{t}_{1}=2 \mathrm{t}_{2}$
Explanation:
D According to Kepler's law of area (Second Law), the areal velocity of a Planet around the sun always remains constant. According to the question- SCD: Area $\left(A_{1}\right)$ swept by planet in time interval $t_{1}$. SAB : Area $\left(\mathrm{A}_{2}\right)$ swept by planet in time interval $\mathrm{t}_{2}$. Given, $\mathrm{A}_{1}=2 \mathrm{~A}_{2}$ As areal velocity is constant. So, $\frac{A_{1}}{t_{1}} =\frac{A_{2}}{t_{2}}$ $t_{1} =t_{2} \cdot \frac{A_{1}}{A_{2}}$ $t_{1} =t_{2} \cdot \frac{2 A_{2}}{A_{2}}$ $t_{1} =2 t_{2}$
