270534
If a and \(b\) are the nearest and farthest distances of a planet from the sun and the planet is revolving in elliptical orbit, then square of the time period of revolution of that planet is proportional to
1 \(a^{3}\)
2 \(b^{3}\)
3 \((a+b)^{3}\)
4 \((a-b)^{3}\)
Explanation:
From Kepler's 3rd law, \(T^{2} \propto r^{3}\)
Gravitation
270535
Let ' \(A\) ' be the area swept by the line joining the earth and the sun during Feb 2007. The area swept by the same line during the first week of that month is
270536
The period of a satellite in an orbit of radius \(R\) is T. Its period of revolution in an orbit of radius \(4 R\) will be
1 \(2 \mathrm{~T}\)
2 \(2 \sqrt{2} T\)
3 \(4 \mathrm{~T}\)
4 \(8 \mathrm{~T}\)
Explanation:
From Kepler's 3rd law, \(T^{2} \propto r^{3}\)
Gravitation
270537
The period of revolution of an earth's satellite close to the surface of earth is 60 minutes. The period of another earth's satellite in an orbit at a distance of three times earth's radius from its surface will be (in minutes)
1 90
2 \(90 \times \sqrt{8}\)
3 270
4 480
Explanation:
From Kepler's 3rd law, \(T^{2} \propto r^{3}\)
Gravitation
270538
If a planet of mass \(m\) is revolving around the sun in a circular orbit of radius \(r\) with time period \(T\), then mass of the sun is
1 \(4 \pi^{2} r^{3} / G T\)
2 \(4 \pi^{2} r^{3} / G T^{2}\)
3 \(4 \pi^{2} r / G T\)
4 \(4 \pi^{2} r^{3} / G^{2} T^{2}\)
Explanation:
From Kepler's 3rd law, \(m r w^{2}=\frac{G M m}{r^{2}} T=2 \pi \sqrt{\frac{r^{3}}{G M}}\)
270534
If a and \(b\) are the nearest and farthest distances of a planet from the sun and the planet is revolving in elliptical orbit, then square of the time period of revolution of that planet is proportional to
1 \(a^{3}\)
2 \(b^{3}\)
3 \((a+b)^{3}\)
4 \((a-b)^{3}\)
Explanation:
From Kepler's 3rd law, \(T^{2} \propto r^{3}\)
Gravitation
270535
Let ' \(A\) ' be the area swept by the line joining the earth and the sun during Feb 2007. The area swept by the same line during the first week of that month is
270536
The period of a satellite in an orbit of radius \(R\) is T. Its period of revolution in an orbit of radius \(4 R\) will be
1 \(2 \mathrm{~T}\)
2 \(2 \sqrt{2} T\)
3 \(4 \mathrm{~T}\)
4 \(8 \mathrm{~T}\)
Explanation:
From Kepler's 3rd law, \(T^{2} \propto r^{3}\)
Gravitation
270537
The period of revolution of an earth's satellite close to the surface of earth is 60 minutes. The period of another earth's satellite in an orbit at a distance of three times earth's radius from its surface will be (in minutes)
1 90
2 \(90 \times \sqrt{8}\)
3 270
4 480
Explanation:
From Kepler's 3rd law, \(T^{2} \propto r^{3}\)
Gravitation
270538
If a planet of mass \(m\) is revolving around the sun in a circular orbit of radius \(r\) with time period \(T\), then mass of the sun is
1 \(4 \pi^{2} r^{3} / G T\)
2 \(4 \pi^{2} r^{3} / G T^{2}\)
3 \(4 \pi^{2} r / G T\)
4 \(4 \pi^{2} r^{3} / G^{2} T^{2}\)
Explanation:
From Kepler's 3rd law, \(m r w^{2}=\frac{G M m}{r^{2}} T=2 \pi \sqrt{\frac{r^{3}}{G M}}\)
270534
If a and \(b\) are the nearest and farthest distances of a planet from the sun and the planet is revolving in elliptical orbit, then square of the time period of revolution of that planet is proportional to
1 \(a^{3}\)
2 \(b^{3}\)
3 \((a+b)^{3}\)
4 \((a-b)^{3}\)
Explanation:
From Kepler's 3rd law, \(T^{2} \propto r^{3}\)
Gravitation
270535
Let ' \(A\) ' be the area swept by the line joining the earth and the sun during Feb 2007. The area swept by the same line during the first week of that month is
270536
The period of a satellite in an orbit of radius \(R\) is T. Its period of revolution in an orbit of radius \(4 R\) will be
1 \(2 \mathrm{~T}\)
2 \(2 \sqrt{2} T\)
3 \(4 \mathrm{~T}\)
4 \(8 \mathrm{~T}\)
Explanation:
From Kepler's 3rd law, \(T^{2} \propto r^{3}\)
Gravitation
270537
The period of revolution of an earth's satellite close to the surface of earth is 60 minutes. The period of another earth's satellite in an orbit at a distance of three times earth's radius from its surface will be (in minutes)
1 90
2 \(90 \times \sqrt{8}\)
3 270
4 480
Explanation:
From Kepler's 3rd law, \(T^{2} \propto r^{3}\)
Gravitation
270538
If a planet of mass \(m\) is revolving around the sun in a circular orbit of radius \(r\) with time period \(T\), then mass of the sun is
1 \(4 \pi^{2} r^{3} / G T\)
2 \(4 \pi^{2} r^{3} / G T^{2}\)
3 \(4 \pi^{2} r / G T\)
4 \(4 \pi^{2} r^{3} / G^{2} T^{2}\)
Explanation:
From Kepler's 3rd law, \(m r w^{2}=\frac{G M m}{r^{2}} T=2 \pi \sqrt{\frac{r^{3}}{G M}}\)
270534
If a and \(b\) are the nearest and farthest distances of a planet from the sun and the planet is revolving in elliptical orbit, then square of the time period of revolution of that planet is proportional to
1 \(a^{3}\)
2 \(b^{3}\)
3 \((a+b)^{3}\)
4 \((a-b)^{3}\)
Explanation:
From Kepler's 3rd law, \(T^{2} \propto r^{3}\)
Gravitation
270535
Let ' \(A\) ' be the area swept by the line joining the earth and the sun during Feb 2007. The area swept by the same line during the first week of that month is
270536
The period of a satellite in an orbit of radius \(R\) is T. Its period of revolution in an orbit of radius \(4 R\) will be
1 \(2 \mathrm{~T}\)
2 \(2 \sqrt{2} T\)
3 \(4 \mathrm{~T}\)
4 \(8 \mathrm{~T}\)
Explanation:
From Kepler's 3rd law, \(T^{2} \propto r^{3}\)
Gravitation
270537
The period of revolution of an earth's satellite close to the surface of earth is 60 minutes. The period of another earth's satellite in an orbit at a distance of three times earth's radius from its surface will be (in minutes)
1 90
2 \(90 \times \sqrt{8}\)
3 270
4 480
Explanation:
From Kepler's 3rd law, \(T^{2} \propto r^{3}\)
Gravitation
270538
If a planet of mass \(m\) is revolving around the sun in a circular orbit of radius \(r\) with time period \(T\), then mass of the sun is
1 \(4 \pi^{2} r^{3} / G T\)
2 \(4 \pi^{2} r^{3} / G T^{2}\)
3 \(4 \pi^{2} r / G T\)
4 \(4 \pi^{2} r^{3} / G^{2} T^{2}\)
Explanation:
From Kepler's 3rd law, \(m r w^{2}=\frac{G M m}{r^{2}} T=2 \pi \sqrt{\frac{r^{3}}{G M}}\)
270534
If a and \(b\) are the nearest and farthest distances of a planet from the sun and the planet is revolving in elliptical orbit, then square of the time period of revolution of that planet is proportional to
1 \(a^{3}\)
2 \(b^{3}\)
3 \((a+b)^{3}\)
4 \((a-b)^{3}\)
Explanation:
From Kepler's 3rd law, \(T^{2} \propto r^{3}\)
Gravitation
270535
Let ' \(A\) ' be the area swept by the line joining the earth and the sun during Feb 2007. The area swept by the same line during the first week of that month is
270536
The period of a satellite in an orbit of radius \(R\) is T. Its period of revolution in an orbit of radius \(4 R\) will be
1 \(2 \mathrm{~T}\)
2 \(2 \sqrt{2} T\)
3 \(4 \mathrm{~T}\)
4 \(8 \mathrm{~T}\)
Explanation:
From Kepler's 3rd law, \(T^{2} \propto r^{3}\)
Gravitation
270537
The period of revolution of an earth's satellite close to the surface of earth is 60 minutes. The period of another earth's satellite in an orbit at a distance of three times earth's radius from its surface will be (in minutes)
1 90
2 \(90 \times \sqrt{8}\)
3 270
4 480
Explanation:
From Kepler's 3rd law, \(T^{2} \propto r^{3}\)
Gravitation
270538
If a planet of mass \(m\) is revolving around the sun in a circular orbit of radius \(r\) with time period \(T\), then mass of the sun is
1 \(4 \pi^{2} r^{3} / G T\)
2 \(4 \pi^{2} r^{3} / G T^{2}\)
3 \(4 \pi^{2} r / G T\)
4 \(4 \pi^{2} r^{3} / G^{2} T^{2}\)
Explanation:
From Kepler's 3rd law, \(m r w^{2}=\frac{G M m}{r^{2}} T=2 \pi \sqrt{\frac{r^{3}}{G M}}\)
