359950
The earth moves around the Sun in an elliptical orbit as shown in the figure. The ratio \(OA/OB = x\). The ratio of the speed of the earth at \(B\) to that at \(A\) is nearly.
1 \(x\)
2 \(x^{2}\)
3 \(x \sqrt{x}\)
4 \(\sqrt{x}\)
Explanation:
Applying conservation of angular momentum at positions \(A\) and \(B\) \(m v_{A} \times O A=m v_{B} \times O B\) Hence, \(\quad \dfrac{v_{B}}{v_{A}}=\dfrac{O A}{O B}=x\)
PHXI08:GRAVITATION
359951
A planet moving around sun sweeps area \(A_{1}\) in 2 days, \(A_{2}\) in 3 days and \(A_{3}\) in 6 days. Then the relation between \(A_{1}, A_{2}\) and \(A_{3}\) is
1 \(3\;{A_1} = 2\;{A_2} = {A_3}\)
2 \(2\;{A_1} = 3\;{A_2} = 6\;{A_3}\)
3 \(3\;{A_1} = 2\;{A_2} = 6\;{A_3}\)
4 \(6\;{A_1} = {34_2} = 2\;{A_3}\)
Explanation:
When a planet revolves around the sun, its area velocity is constant \(\therefore \frac{{{A_1}}}{{{t_1}}} = \frac{{{A_2}}}{{{t_2}}} = \frac{{{A_3}}}{{{t_3}}}\) \(\frac{{{A_1}}}{2} = \frac{{{A_2}}}{3} = \frac{{{A_3}}}{6} \Rightarrow 3\;{A_1} = 2\;{A_2} = {A_3}\)
KCET - 2012
PHXI08:GRAVITATION
359952
A planet is revolving around the sun as shown in elliptical path. The correct option is
1 The time taken in travelling \(CDA\) is greater than that for \(ABC\)
2 The time taken in travelling \(DAB\) is less than that for \(BCD\)
3 The time taken in travelling \(DAB\) is greater than that for \(BCD\)
4 The time taken in travelling \(CDA\) is less than that for \(ABC\)
Explanation:
During path \(D A B\) planet is nearer to sun as comparing with path \(BCD\). From conservation of angular momentum \(v r=\) constant. So time taken in travelling \(DAB\) is less than that for \(BCD\) because velocity of planet will be more in region \(DAB\)
PHXI08:GRAVITATION
359953
A planet of mass \(m\) moves along an elliptical path around sun so that perihelion and aphelion distances are \(r_{1}\) and \(r_{2}\). Find the angular momentum of the planet.
1 \(L=m \sqrt{2 G M_{S}\left(r_{1}+r_{2}\right)}\)
3 \(L=m \sqrt{\dfrac{2 G M_{S} r_{1} r_{2}}{r_{1}+r_{2}}}\)
4 \(L=m \sqrt{G M_{S}\left(r_{1}+r_{2}\right)}\)
Explanation:
From conservation of angular momentum \(m v_{1} r_{1}=m v_{2} r_{2}\) Using energy conservation \(\begin{aligned}& \dfrac{-G m M_{S}}{r_{1}}+\dfrac{m v_{1}^{2}}{2}=\dfrac{-G m M_{S}}{r_{2}}+\dfrac{m v_{2}^{2}}{2} \\& \Rightarrow \dfrac{-G M_{S}}{r_{1}}+\dfrac{G M_{S}}{r_{2}}=\dfrac{v_{2}^{2}}{2}-\dfrac{v_{2}^{2}}{2}\left(\dfrac{r_{2}}{r_{1}}\right)^{2} \\& \Rightarrow v_{2}=\sqrt{\dfrac{2 G M_{S} r_{1}}{r_{2}\left(r_{1}+r_{2}\right)}}\end{aligned}\) and \(L=m v_{2} r_{2}=m \sqrt{\dfrac{2 G M_{S} r_{1} r_{2}}{r_{1}+r_{2}}}\)
359950
The earth moves around the Sun in an elliptical orbit as shown in the figure. The ratio \(OA/OB = x\). The ratio of the speed of the earth at \(B\) to that at \(A\) is nearly.
1 \(x\)
2 \(x^{2}\)
3 \(x \sqrt{x}\)
4 \(\sqrt{x}\)
Explanation:
Applying conservation of angular momentum at positions \(A\) and \(B\) \(m v_{A} \times O A=m v_{B} \times O B\) Hence, \(\quad \dfrac{v_{B}}{v_{A}}=\dfrac{O A}{O B}=x\)
PHXI08:GRAVITATION
359951
A planet moving around sun sweeps area \(A_{1}\) in 2 days, \(A_{2}\) in 3 days and \(A_{3}\) in 6 days. Then the relation between \(A_{1}, A_{2}\) and \(A_{3}\) is
1 \(3\;{A_1} = 2\;{A_2} = {A_3}\)
2 \(2\;{A_1} = 3\;{A_2} = 6\;{A_3}\)
3 \(3\;{A_1} = 2\;{A_2} = 6\;{A_3}\)
4 \(6\;{A_1} = {34_2} = 2\;{A_3}\)
Explanation:
When a planet revolves around the sun, its area velocity is constant \(\therefore \frac{{{A_1}}}{{{t_1}}} = \frac{{{A_2}}}{{{t_2}}} = \frac{{{A_3}}}{{{t_3}}}\) \(\frac{{{A_1}}}{2} = \frac{{{A_2}}}{3} = \frac{{{A_3}}}{6} \Rightarrow 3\;{A_1} = 2\;{A_2} = {A_3}\)
KCET - 2012
PHXI08:GRAVITATION
359952
A planet is revolving around the sun as shown in elliptical path. The correct option is
1 The time taken in travelling \(CDA\) is greater than that for \(ABC\)
2 The time taken in travelling \(DAB\) is less than that for \(BCD\)
3 The time taken in travelling \(DAB\) is greater than that for \(BCD\)
4 The time taken in travelling \(CDA\) is less than that for \(ABC\)
Explanation:
During path \(D A B\) planet is nearer to sun as comparing with path \(BCD\). From conservation of angular momentum \(v r=\) constant. So time taken in travelling \(DAB\) is less than that for \(BCD\) because velocity of planet will be more in region \(DAB\)
PHXI08:GRAVITATION
359953
A planet of mass \(m\) moves along an elliptical path around sun so that perihelion and aphelion distances are \(r_{1}\) and \(r_{2}\). Find the angular momentum of the planet.
1 \(L=m \sqrt{2 G M_{S}\left(r_{1}+r_{2}\right)}\)
3 \(L=m \sqrt{\dfrac{2 G M_{S} r_{1} r_{2}}{r_{1}+r_{2}}}\)
4 \(L=m \sqrt{G M_{S}\left(r_{1}+r_{2}\right)}\)
Explanation:
From conservation of angular momentum \(m v_{1} r_{1}=m v_{2} r_{2}\) Using energy conservation \(\begin{aligned}& \dfrac{-G m M_{S}}{r_{1}}+\dfrac{m v_{1}^{2}}{2}=\dfrac{-G m M_{S}}{r_{2}}+\dfrac{m v_{2}^{2}}{2} \\& \Rightarrow \dfrac{-G M_{S}}{r_{1}}+\dfrac{G M_{S}}{r_{2}}=\dfrac{v_{2}^{2}}{2}-\dfrac{v_{2}^{2}}{2}\left(\dfrac{r_{2}}{r_{1}}\right)^{2} \\& \Rightarrow v_{2}=\sqrt{\dfrac{2 G M_{S} r_{1}}{r_{2}\left(r_{1}+r_{2}\right)}}\end{aligned}\) and \(L=m v_{2} r_{2}=m \sqrt{\dfrac{2 G M_{S} r_{1} r_{2}}{r_{1}+r_{2}}}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI08:GRAVITATION
359950
The earth moves around the Sun in an elliptical orbit as shown in the figure. The ratio \(OA/OB = x\). The ratio of the speed of the earth at \(B\) to that at \(A\) is nearly.
1 \(x\)
2 \(x^{2}\)
3 \(x \sqrt{x}\)
4 \(\sqrt{x}\)
Explanation:
Applying conservation of angular momentum at positions \(A\) and \(B\) \(m v_{A} \times O A=m v_{B} \times O B\) Hence, \(\quad \dfrac{v_{B}}{v_{A}}=\dfrac{O A}{O B}=x\)
PHXI08:GRAVITATION
359951
A planet moving around sun sweeps area \(A_{1}\) in 2 days, \(A_{2}\) in 3 days and \(A_{3}\) in 6 days. Then the relation between \(A_{1}, A_{2}\) and \(A_{3}\) is
1 \(3\;{A_1} = 2\;{A_2} = {A_3}\)
2 \(2\;{A_1} = 3\;{A_2} = 6\;{A_3}\)
3 \(3\;{A_1} = 2\;{A_2} = 6\;{A_3}\)
4 \(6\;{A_1} = {34_2} = 2\;{A_3}\)
Explanation:
When a planet revolves around the sun, its area velocity is constant \(\therefore \frac{{{A_1}}}{{{t_1}}} = \frac{{{A_2}}}{{{t_2}}} = \frac{{{A_3}}}{{{t_3}}}\) \(\frac{{{A_1}}}{2} = \frac{{{A_2}}}{3} = \frac{{{A_3}}}{6} \Rightarrow 3\;{A_1} = 2\;{A_2} = {A_3}\)
KCET - 2012
PHXI08:GRAVITATION
359952
A planet is revolving around the sun as shown in elliptical path. The correct option is
1 The time taken in travelling \(CDA\) is greater than that for \(ABC\)
2 The time taken in travelling \(DAB\) is less than that for \(BCD\)
3 The time taken in travelling \(DAB\) is greater than that for \(BCD\)
4 The time taken in travelling \(CDA\) is less than that for \(ABC\)
Explanation:
During path \(D A B\) planet is nearer to sun as comparing with path \(BCD\). From conservation of angular momentum \(v r=\) constant. So time taken in travelling \(DAB\) is less than that for \(BCD\) because velocity of planet will be more in region \(DAB\)
PHXI08:GRAVITATION
359953
A planet of mass \(m\) moves along an elliptical path around sun so that perihelion and aphelion distances are \(r_{1}\) and \(r_{2}\). Find the angular momentum of the planet.
1 \(L=m \sqrt{2 G M_{S}\left(r_{1}+r_{2}\right)}\)
3 \(L=m \sqrt{\dfrac{2 G M_{S} r_{1} r_{2}}{r_{1}+r_{2}}}\)
4 \(L=m \sqrt{G M_{S}\left(r_{1}+r_{2}\right)}\)
Explanation:
From conservation of angular momentum \(m v_{1} r_{1}=m v_{2} r_{2}\) Using energy conservation \(\begin{aligned}& \dfrac{-G m M_{S}}{r_{1}}+\dfrac{m v_{1}^{2}}{2}=\dfrac{-G m M_{S}}{r_{2}}+\dfrac{m v_{2}^{2}}{2} \\& \Rightarrow \dfrac{-G M_{S}}{r_{1}}+\dfrac{G M_{S}}{r_{2}}=\dfrac{v_{2}^{2}}{2}-\dfrac{v_{2}^{2}}{2}\left(\dfrac{r_{2}}{r_{1}}\right)^{2} \\& \Rightarrow v_{2}=\sqrt{\dfrac{2 G M_{S} r_{1}}{r_{2}\left(r_{1}+r_{2}\right)}}\end{aligned}\) and \(L=m v_{2} r_{2}=m \sqrt{\dfrac{2 G M_{S} r_{1} r_{2}}{r_{1}+r_{2}}}\)
359950
The earth moves around the Sun in an elliptical orbit as shown in the figure. The ratio \(OA/OB = x\). The ratio of the speed of the earth at \(B\) to that at \(A\) is nearly.
1 \(x\)
2 \(x^{2}\)
3 \(x \sqrt{x}\)
4 \(\sqrt{x}\)
Explanation:
Applying conservation of angular momentum at positions \(A\) and \(B\) \(m v_{A} \times O A=m v_{B} \times O B\) Hence, \(\quad \dfrac{v_{B}}{v_{A}}=\dfrac{O A}{O B}=x\)
PHXI08:GRAVITATION
359951
A planet moving around sun sweeps area \(A_{1}\) in 2 days, \(A_{2}\) in 3 days and \(A_{3}\) in 6 days. Then the relation between \(A_{1}, A_{2}\) and \(A_{3}\) is
1 \(3\;{A_1} = 2\;{A_2} = {A_3}\)
2 \(2\;{A_1} = 3\;{A_2} = 6\;{A_3}\)
3 \(3\;{A_1} = 2\;{A_2} = 6\;{A_3}\)
4 \(6\;{A_1} = {34_2} = 2\;{A_3}\)
Explanation:
When a planet revolves around the sun, its area velocity is constant \(\therefore \frac{{{A_1}}}{{{t_1}}} = \frac{{{A_2}}}{{{t_2}}} = \frac{{{A_3}}}{{{t_3}}}\) \(\frac{{{A_1}}}{2} = \frac{{{A_2}}}{3} = \frac{{{A_3}}}{6} \Rightarrow 3\;{A_1} = 2\;{A_2} = {A_3}\)
KCET - 2012
PHXI08:GRAVITATION
359952
A planet is revolving around the sun as shown in elliptical path. The correct option is
1 The time taken in travelling \(CDA\) is greater than that for \(ABC\)
2 The time taken in travelling \(DAB\) is less than that for \(BCD\)
3 The time taken in travelling \(DAB\) is greater than that for \(BCD\)
4 The time taken in travelling \(CDA\) is less than that for \(ABC\)
Explanation:
During path \(D A B\) planet is nearer to sun as comparing with path \(BCD\). From conservation of angular momentum \(v r=\) constant. So time taken in travelling \(DAB\) is less than that for \(BCD\) because velocity of planet will be more in region \(DAB\)
PHXI08:GRAVITATION
359953
A planet of mass \(m\) moves along an elliptical path around sun so that perihelion and aphelion distances are \(r_{1}\) and \(r_{2}\). Find the angular momentum of the planet.
1 \(L=m \sqrt{2 G M_{S}\left(r_{1}+r_{2}\right)}\)
3 \(L=m \sqrt{\dfrac{2 G M_{S} r_{1} r_{2}}{r_{1}+r_{2}}}\)
4 \(L=m \sqrt{G M_{S}\left(r_{1}+r_{2}\right)}\)
Explanation:
From conservation of angular momentum \(m v_{1} r_{1}=m v_{2} r_{2}\) Using energy conservation \(\begin{aligned}& \dfrac{-G m M_{S}}{r_{1}}+\dfrac{m v_{1}^{2}}{2}=\dfrac{-G m M_{S}}{r_{2}}+\dfrac{m v_{2}^{2}}{2} \\& \Rightarrow \dfrac{-G M_{S}}{r_{1}}+\dfrac{G M_{S}}{r_{2}}=\dfrac{v_{2}^{2}}{2}-\dfrac{v_{2}^{2}}{2}\left(\dfrac{r_{2}}{r_{1}}\right)^{2} \\& \Rightarrow v_{2}=\sqrt{\dfrac{2 G M_{S} r_{1}}{r_{2}\left(r_{1}+r_{2}\right)}}\end{aligned}\) and \(L=m v_{2} r_{2}=m \sqrt{\dfrac{2 G M_{S} r_{1} r_{2}}{r_{1}+r_{2}}}\)
