270539
The period of revolution of a planet around the sun in a circular orbit is same as that of period of similar planet revolving around a star of twice the radius of first orbit and ' \(M\) ' is the mass of the sun and mass of star is
1 \(2 \mathrm{M}\)
2 \(4 \mathrm{M}\)
3 \(8 \mathrm{M}\)
4 \(16 \mathrm{M}\)
Explanation:
From Kepler's 3rd law, \(m r w^{2}=\frac{G M m}{r^{2}} T=2 \pi \sqrt{\frac{r^{3}}{G M}}\)
Gravitation
270540
A planet moves around the sun in elliptical orbit. When earth is closest from the sun, it is at a distance \(r\) having a speed \(v\). When it is at a distance \(4 r\) from the sun its speed is
1 \(4 \mathrm{v}\)
2 \(\frac{v}{4}\)
3 \(2 \mathrm{v}\)
4 \(\frac{v}{2}\)
Explanation:
From Kepler's 2nd law, \(\frac{1}{2} r \mathrm{v}=\) Constant
Gravitation
270541
A planet of mass ' \(m\) ' is in an elliptical orbit about the sun \((m \ll M)\) with an orbital time period ' \(T\) '. If ' \(A\) ' be the area of the orbit then its angular momentum is
1 \(\frac{2 m A}{T}\)
2 \(m A T\)
3 \(\frac{m A}{2 T}\)
4 \(2 m A T\)
Explanation:
\(\frac{d A}{d t}=\frac{L}{2 m}\)
Gravitation
270589
If the Earth shrinks such that its density becomes 8 times to the present value, then new duration of the day in hours will be
1 24
2 12
3 6
4 3
Explanation:
Given \(m_{1}=m_{2} \Rightarrow V_{1} d_{1}=V_{2} d_{2}\)
\(\Rightarrow R_{1}^{3} d_{1}=R_{2}^{3} d_{2} \Rightarrow R_{1}=2 R_{2}\)
From law of conservation of angular momentum
\(I_{1} \omega_{1}=I_{2} \omega_{2} \Rightarrow \frac{T_{2}}{T_{1}}=\square \frac{R_{2}}{\square} \frac{R_{1}}{} \square\)
270539
The period of revolution of a planet around the sun in a circular orbit is same as that of period of similar planet revolving around a star of twice the radius of first orbit and ' \(M\) ' is the mass of the sun and mass of star is
1 \(2 \mathrm{M}\)
2 \(4 \mathrm{M}\)
3 \(8 \mathrm{M}\)
4 \(16 \mathrm{M}\)
Explanation:
From Kepler's 3rd law, \(m r w^{2}=\frac{G M m}{r^{2}} T=2 \pi \sqrt{\frac{r^{3}}{G M}}\)
Gravitation
270540
A planet moves around the sun in elliptical orbit. When earth is closest from the sun, it is at a distance \(r\) having a speed \(v\). When it is at a distance \(4 r\) from the sun its speed is
1 \(4 \mathrm{v}\)
2 \(\frac{v}{4}\)
3 \(2 \mathrm{v}\)
4 \(\frac{v}{2}\)
Explanation:
From Kepler's 2nd law, \(\frac{1}{2} r \mathrm{v}=\) Constant
Gravitation
270541
A planet of mass ' \(m\) ' is in an elliptical orbit about the sun \((m \ll M)\) with an orbital time period ' \(T\) '. If ' \(A\) ' be the area of the orbit then its angular momentum is
1 \(\frac{2 m A}{T}\)
2 \(m A T\)
3 \(\frac{m A}{2 T}\)
4 \(2 m A T\)
Explanation:
\(\frac{d A}{d t}=\frac{L}{2 m}\)
Gravitation
270589
If the Earth shrinks such that its density becomes 8 times to the present value, then new duration of the day in hours will be
1 24
2 12
3 6
4 3
Explanation:
Given \(m_{1}=m_{2} \Rightarrow V_{1} d_{1}=V_{2} d_{2}\)
\(\Rightarrow R_{1}^{3} d_{1}=R_{2}^{3} d_{2} \Rightarrow R_{1}=2 R_{2}\)
From law of conservation of angular momentum
\(I_{1} \omega_{1}=I_{2} \omega_{2} \Rightarrow \frac{T_{2}}{T_{1}}=\square \frac{R_{2}}{\square} \frac{R_{1}}{} \square\)
270539
The period of revolution of a planet around the sun in a circular orbit is same as that of period of similar planet revolving around a star of twice the radius of first orbit and ' \(M\) ' is the mass of the sun and mass of star is
1 \(2 \mathrm{M}\)
2 \(4 \mathrm{M}\)
3 \(8 \mathrm{M}\)
4 \(16 \mathrm{M}\)
Explanation:
From Kepler's 3rd law, \(m r w^{2}=\frac{G M m}{r^{2}} T=2 \pi \sqrt{\frac{r^{3}}{G M}}\)
Gravitation
270540
A planet moves around the sun in elliptical orbit. When earth is closest from the sun, it is at a distance \(r\) having a speed \(v\). When it is at a distance \(4 r\) from the sun its speed is
1 \(4 \mathrm{v}\)
2 \(\frac{v}{4}\)
3 \(2 \mathrm{v}\)
4 \(\frac{v}{2}\)
Explanation:
From Kepler's 2nd law, \(\frac{1}{2} r \mathrm{v}=\) Constant
Gravitation
270541
A planet of mass ' \(m\) ' is in an elliptical orbit about the sun \((m \ll M)\) with an orbital time period ' \(T\) '. If ' \(A\) ' be the area of the orbit then its angular momentum is
1 \(\frac{2 m A}{T}\)
2 \(m A T\)
3 \(\frac{m A}{2 T}\)
4 \(2 m A T\)
Explanation:
\(\frac{d A}{d t}=\frac{L}{2 m}\)
Gravitation
270589
If the Earth shrinks such that its density becomes 8 times to the present value, then new duration of the day in hours will be
1 24
2 12
3 6
4 3
Explanation:
Given \(m_{1}=m_{2} \Rightarrow V_{1} d_{1}=V_{2} d_{2}\)
\(\Rightarrow R_{1}^{3} d_{1}=R_{2}^{3} d_{2} \Rightarrow R_{1}=2 R_{2}\)
From law of conservation of angular momentum
\(I_{1} \omega_{1}=I_{2} \omega_{2} \Rightarrow \frac{T_{2}}{T_{1}}=\square \frac{R_{2}}{\square} \frac{R_{1}}{} \square\)
270539
The period of revolution of a planet around the sun in a circular orbit is same as that of period of similar planet revolving around a star of twice the radius of first orbit and ' \(M\) ' is the mass of the sun and mass of star is
1 \(2 \mathrm{M}\)
2 \(4 \mathrm{M}\)
3 \(8 \mathrm{M}\)
4 \(16 \mathrm{M}\)
Explanation:
From Kepler's 3rd law, \(m r w^{2}=\frac{G M m}{r^{2}} T=2 \pi \sqrt{\frac{r^{3}}{G M}}\)
Gravitation
270540
A planet moves around the sun in elliptical orbit. When earth is closest from the sun, it is at a distance \(r\) having a speed \(v\). When it is at a distance \(4 r\) from the sun its speed is
1 \(4 \mathrm{v}\)
2 \(\frac{v}{4}\)
3 \(2 \mathrm{v}\)
4 \(\frac{v}{2}\)
Explanation:
From Kepler's 2nd law, \(\frac{1}{2} r \mathrm{v}=\) Constant
Gravitation
270541
A planet of mass ' \(m\) ' is in an elliptical orbit about the sun \((m \ll M)\) with an orbital time period ' \(T\) '. If ' \(A\) ' be the area of the orbit then its angular momentum is
1 \(\frac{2 m A}{T}\)
2 \(m A T\)
3 \(\frac{m A}{2 T}\)
4 \(2 m A T\)
Explanation:
\(\frac{d A}{d t}=\frac{L}{2 m}\)
Gravitation
270589
If the Earth shrinks such that its density becomes 8 times to the present value, then new duration of the day in hours will be
1 24
2 12
3 6
4 3
Explanation:
Given \(m_{1}=m_{2} \Rightarrow V_{1} d_{1}=V_{2} d_{2}\)
\(\Rightarrow R_{1}^{3} d_{1}=R_{2}^{3} d_{2} \Rightarrow R_{1}=2 R_{2}\)
From law of conservation of angular momentum
\(I_{1} \omega_{1}=I_{2} \omega_{2} \Rightarrow \frac{T_{2}}{T_{1}}=\square \frac{R_{2}}{\square} \frac{R_{1}}{} \square\)
