138644
The distance between Sun and Earth is R. The duration of year if the distance between Sun and Earth becomes $3 R$ will be:
1 $\sqrt{3}$ years
2 3 years
3 8 years
4 $3 \sqrt{3}$ years
Explanation:
D Given, $\mathrm{T}=1$ year According to Kepler's third law $\mathrm{T}^{2} \propto \mathrm{r}^{3}$ If the distance between Sun and Earth is R $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ If the distance between Sun and Earth is $3 R$ $\left(\mathrm{T}^{\prime}\right)^{2} \propto(3 \mathrm{R})^{3}$ Dividing equation (ii) by (i), we get - $\left(\frac{\mathrm{T}^{\prime}}{\mathrm{T}}\right)^{2}=\left(\frac{3 \mathrm{R}}{\mathrm{R}}\right)^{3}$ $\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=(3)^{3 / 2}$ $\mathrm{~T}^{\prime}=3 \sqrt{3} \mathrm{~T}$ So, $T^{\prime}=3 \sqrt{3}$ years
JEE Main-24.06.2022
Gravitation
138646
A satellite is to be placed in equatorial geostationary orbit around the Earth for communication purpose. The height of such satellite is $\left(M_{E}=6 \times 10^{24} \mathrm{~kg}, R_{E}=6400 \mathrm{~km}\right)$
1 $3.57 \times 10^{8} \mathrm{~m}$
2 $3.57 \times 10^{7} \mathrm{~m}$
3 $3.57 \times 10^{5} \mathrm{~m}$
4 $3.57 \times 10^{6} \mathrm{~m}$
Explanation:
B As we know, Time period of satellite is given by, $\mathrm{T}=\frac{2 \pi\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)}{\mathrm{v}}$ Orbital velocity of satellite is given by, $v=\sqrt{\frac{\mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}+\mathrm{h}}}$ Putting the value of $\mathrm{v}$ in equation (i), time period of satellite, $\mathrm{T}=\frac{2 \pi\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)}{\sqrt{\frac{\mathrm{GM}_{\mathrm{E}}}{\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)}}}$ $=\frac{2 \pi\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)^{\frac{3}{2}}}{\sqrt{\mathrm{GM}_{\mathrm{E}}}}$ Squaring both sides, we get - $\mathrm{T}^{2}=\frac{4 \pi^{2}\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)^{3}}{\mathrm{GM}_{\mathrm{E}}}$ $\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)^{3}=\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{T}^{2}}{4 \pi^{2}}$ $\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)=\left(\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{T}^{2}}{4 \pi^{2}}\right)^{1 / 3}$ $\mathrm{~h}=\left(\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{T}^{2}}{4 \pi^{2}}\right)^{1 / 3}-\mathrm{R}_{\mathrm{E}}$ Here, $\quad M_{E}=6 \times 10^{24} \mathrm{~kg}$ $\mathrm{R}_{\mathrm{e}}=6400 \mathrm{~km}=6400 \times 10^{3} \mathrm{~m}$ $=6.4 \times 10^{6} \mathrm{~m}$ $\mathrm{T}=24 \mathrm{~h}=24 \times 60 \times 60 \mathrm{~s}=86400 \mathrm{~s}$ $\mathrm{G}=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}$ On substituting the given values in equation (ii), we get $\mathrm{h}=\left(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times(86400)^{2}}{4 \times(3.14)^{2}}\right)^{1 / 3}-6.4 \times 10^{6}$ $\mathrm{~h}=4.21 \times 10^{7}-6.4 \times 10^{6}$ $=3.57 \times 10^{7} \mathrm{~m}$
AP EAMCET-24.04.2018
Gravitation
138647
A planet is revolving round the sun of mass ' $M$ ' in an elliptical orbit with semi-major axis ' $a$ '. The speed of the planer when it is a distance ' $r$ ' from the sun is. (G - Universal gravitational constant)
A In elliptical orbit, total energy. $\mathrm{E}=\frac{-\mathrm{GMm}}{2 \mathrm{a}} \quad(\mathrm{m}=\text { mass of planet })$ From conservation of mechanical energy $\text { K.E }+ \text { P.E. }=E$ $\frac{1}{2} m v^{2}-\frac{G M m}{r}=\frac{-G M m}{2 a}$ $\frac{1}{2} m v^{2}=\frac{G M m}{r}-\frac{G M m}{2 a}$ $\frac{1}{2} v^{2}=G M\left(\frac{1}{r}-\frac{1}{2 a}\right)$ $v=\sqrt{G M\left(\frac{2}{r}-\frac{1}{a}\right)}$
AP EAMCET (Medical)-24.04.2019
Gravitation
138648
The time period of moon around the earth is 28 days. If the mass of the earth is doubled, without any change in the distance of the moon from the earth. New time period of revolution of the moon is
1 $28 \sqrt{2}$ days
2 7 days
3 $14 \sqrt{2}$ days
4 14 days
Explanation:
C Given, $\mathrm{T}=28$ days We know, $\mathrm{T}^{2}=\frac{4 \pi^{2}}{\mathrm{GM}} \mathrm{r}^{3}$ When the mass of the earth doubled, $\mathrm{T}^{\prime 2}=\frac{4 \pi^{2}}{\mathrm{G} \cdot 2 \mathrm{M}} \mathrm{r}^{3}$ Dividing equation (ii) by (i), we get - $\therefore \quad \quad \frac{\mathrm{T}^{\prime 2}}{\mathrm{~T}^{2}} =\frac{\frac{4 \pi^{2}}{\mathrm{GM}} \cdot \frac{\mathrm{r}^{3}}{2}}{\frac{4 \pi^{2}}{\mathrm{GM}} \cdot \mathrm{r}^{3}}$ $\mathrm{~T}^{\prime 2} =\mathrm{T}^{2} / 2$ $\mathrm{~T}^{\prime} =\frac{\mathrm{T}}{\sqrt{2}}$ $\mathrm{~T}^{\prime} =\frac{28}{\sqrt{2}}$ $\mathrm{~T}^{\prime} =14 \sqrt{2}$
AP EAMCET-24.04.2019
Gravitation
138649
A number of planets are revolving around the Sun. Time period is ' $T$ ' and average orbital radius of a planet is ' $R$ '. A graph is drawn between $\log T$ on the $\mathrm{Y}$-axis and $\log \mathrm{R}$ on the $\mathrm{X}$-axis with the origin at $(0,0)$. The graph is a
1 Straight line with slope $\frac{3}{2}$ and passing through origin
2 Straight line with slope $\frac{3}{2}$ and not passing through origin
3 Parabola
4 Ellipse
Explanation:
B Using Kepler's $3^{\text {rd }}$ law, $T^{2} \propto R^{3}$ $T \propto R^{3 / 2}$ $T=K R^{3 / 2}$ Taking $\log$ of both side $\log \mathrm{T}=\log \left(\mathrm{KR}^{3 / 2}\right)$ $\log \mathrm{T}=\log \mathrm{K}+\frac{3}{2} \log \mathrm{R}$ $\log \mathrm{T}=\frac{3}{2} \log \mathrm{R}+\log \mathrm{K}$
138644
The distance between Sun and Earth is R. The duration of year if the distance between Sun and Earth becomes $3 R$ will be:
1 $\sqrt{3}$ years
2 3 years
3 8 years
4 $3 \sqrt{3}$ years
Explanation:
D Given, $\mathrm{T}=1$ year According to Kepler's third law $\mathrm{T}^{2} \propto \mathrm{r}^{3}$ If the distance between Sun and Earth is R $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ If the distance between Sun and Earth is $3 R$ $\left(\mathrm{T}^{\prime}\right)^{2} \propto(3 \mathrm{R})^{3}$ Dividing equation (ii) by (i), we get - $\left(\frac{\mathrm{T}^{\prime}}{\mathrm{T}}\right)^{2}=\left(\frac{3 \mathrm{R}}{\mathrm{R}}\right)^{3}$ $\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=(3)^{3 / 2}$ $\mathrm{~T}^{\prime}=3 \sqrt{3} \mathrm{~T}$ So, $T^{\prime}=3 \sqrt{3}$ years
JEE Main-24.06.2022
Gravitation
138646
A satellite is to be placed in equatorial geostationary orbit around the Earth for communication purpose. The height of such satellite is $\left(M_{E}=6 \times 10^{24} \mathrm{~kg}, R_{E}=6400 \mathrm{~km}\right)$
1 $3.57 \times 10^{8} \mathrm{~m}$
2 $3.57 \times 10^{7} \mathrm{~m}$
3 $3.57 \times 10^{5} \mathrm{~m}$
4 $3.57 \times 10^{6} \mathrm{~m}$
Explanation:
B As we know, Time period of satellite is given by, $\mathrm{T}=\frac{2 \pi\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)}{\mathrm{v}}$ Orbital velocity of satellite is given by, $v=\sqrt{\frac{\mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}+\mathrm{h}}}$ Putting the value of $\mathrm{v}$ in equation (i), time period of satellite, $\mathrm{T}=\frac{2 \pi\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)}{\sqrt{\frac{\mathrm{GM}_{\mathrm{E}}}{\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)}}}$ $=\frac{2 \pi\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)^{\frac{3}{2}}}{\sqrt{\mathrm{GM}_{\mathrm{E}}}}$ Squaring both sides, we get - $\mathrm{T}^{2}=\frac{4 \pi^{2}\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)^{3}}{\mathrm{GM}_{\mathrm{E}}}$ $\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)^{3}=\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{T}^{2}}{4 \pi^{2}}$ $\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)=\left(\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{T}^{2}}{4 \pi^{2}}\right)^{1 / 3}$ $\mathrm{~h}=\left(\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{T}^{2}}{4 \pi^{2}}\right)^{1 / 3}-\mathrm{R}_{\mathrm{E}}$ Here, $\quad M_{E}=6 \times 10^{24} \mathrm{~kg}$ $\mathrm{R}_{\mathrm{e}}=6400 \mathrm{~km}=6400 \times 10^{3} \mathrm{~m}$ $=6.4 \times 10^{6} \mathrm{~m}$ $\mathrm{T}=24 \mathrm{~h}=24 \times 60 \times 60 \mathrm{~s}=86400 \mathrm{~s}$ $\mathrm{G}=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}$ On substituting the given values in equation (ii), we get $\mathrm{h}=\left(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times(86400)^{2}}{4 \times(3.14)^{2}}\right)^{1 / 3}-6.4 \times 10^{6}$ $\mathrm{~h}=4.21 \times 10^{7}-6.4 \times 10^{6}$ $=3.57 \times 10^{7} \mathrm{~m}$
AP EAMCET-24.04.2018
Gravitation
138647
A planet is revolving round the sun of mass ' $M$ ' in an elliptical orbit with semi-major axis ' $a$ '. The speed of the planer when it is a distance ' $r$ ' from the sun is. (G - Universal gravitational constant)
A In elliptical orbit, total energy. $\mathrm{E}=\frac{-\mathrm{GMm}}{2 \mathrm{a}} \quad(\mathrm{m}=\text { mass of planet })$ From conservation of mechanical energy $\text { K.E }+ \text { P.E. }=E$ $\frac{1}{2} m v^{2}-\frac{G M m}{r}=\frac{-G M m}{2 a}$ $\frac{1}{2} m v^{2}=\frac{G M m}{r}-\frac{G M m}{2 a}$ $\frac{1}{2} v^{2}=G M\left(\frac{1}{r}-\frac{1}{2 a}\right)$ $v=\sqrt{G M\left(\frac{2}{r}-\frac{1}{a}\right)}$
AP EAMCET (Medical)-24.04.2019
Gravitation
138648
The time period of moon around the earth is 28 days. If the mass of the earth is doubled, without any change in the distance of the moon from the earth. New time period of revolution of the moon is
1 $28 \sqrt{2}$ days
2 7 days
3 $14 \sqrt{2}$ days
4 14 days
Explanation:
C Given, $\mathrm{T}=28$ days We know, $\mathrm{T}^{2}=\frac{4 \pi^{2}}{\mathrm{GM}} \mathrm{r}^{3}$ When the mass of the earth doubled, $\mathrm{T}^{\prime 2}=\frac{4 \pi^{2}}{\mathrm{G} \cdot 2 \mathrm{M}} \mathrm{r}^{3}$ Dividing equation (ii) by (i), we get - $\therefore \quad \quad \frac{\mathrm{T}^{\prime 2}}{\mathrm{~T}^{2}} =\frac{\frac{4 \pi^{2}}{\mathrm{GM}} \cdot \frac{\mathrm{r}^{3}}{2}}{\frac{4 \pi^{2}}{\mathrm{GM}} \cdot \mathrm{r}^{3}}$ $\mathrm{~T}^{\prime 2} =\mathrm{T}^{2} / 2$ $\mathrm{~T}^{\prime} =\frac{\mathrm{T}}{\sqrt{2}}$ $\mathrm{~T}^{\prime} =\frac{28}{\sqrt{2}}$ $\mathrm{~T}^{\prime} =14 \sqrt{2}$
AP EAMCET-24.04.2019
Gravitation
138649
A number of planets are revolving around the Sun. Time period is ' $T$ ' and average orbital radius of a planet is ' $R$ '. A graph is drawn between $\log T$ on the $\mathrm{Y}$-axis and $\log \mathrm{R}$ on the $\mathrm{X}$-axis with the origin at $(0,0)$. The graph is a
1 Straight line with slope $\frac{3}{2}$ and passing through origin
2 Straight line with slope $\frac{3}{2}$ and not passing through origin
3 Parabola
4 Ellipse
Explanation:
B Using Kepler's $3^{\text {rd }}$ law, $T^{2} \propto R^{3}$ $T \propto R^{3 / 2}$ $T=K R^{3 / 2}$ Taking $\log$ of both side $\log \mathrm{T}=\log \left(\mathrm{KR}^{3 / 2}\right)$ $\log \mathrm{T}=\log \mathrm{K}+\frac{3}{2} \log \mathrm{R}$ $\log \mathrm{T}=\frac{3}{2} \log \mathrm{R}+\log \mathrm{K}$
138644
The distance between Sun and Earth is R. The duration of year if the distance between Sun and Earth becomes $3 R$ will be:
1 $\sqrt{3}$ years
2 3 years
3 8 years
4 $3 \sqrt{3}$ years
Explanation:
D Given, $\mathrm{T}=1$ year According to Kepler's third law $\mathrm{T}^{2} \propto \mathrm{r}^{3}$ If the distance between Sun and Earth is R $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ If the distance between Sun and Earth is $3 R$ $\left(\mathrm{T}^{\prime}\right)^{2} \propto(3 \mathrm{R})^{3}$ Dividing equation (ii) by (i), we get - $\left(\frac{\mathrm{T}^{\prime}}{\mathrm{T}}\right)^{2}=\left(\frac{3 \mathrm{R}}{\mathrm{R}}\right)^{3}$ $\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=(3)^{3 / 2}$ $\mathrm{~T}^{\prime}=3 \sqrt{3} \mathrm{~T}$ So, $T^{\prime}=3 \sqrt{3}$ years
JEE Main-24.06.2022
Gravitation
138646
A satellite is to be placed in equatorial geostationary orbit around the Earth for communication purpose. The height of such satellite is $\left(M_{E}=6 \times 10^{24} \mathrm{~kg}, R_{E}=6400 \mathrm{~km}\right)$
1 $3.57 \times 10^{8} \mathrm{~m}$
2 $3.57 \times 10^{7} \mathrm{~m}$
3 $3.57 \times 10^{5} \mathrm{~m}$
4 $3.57 \times 10^{6} \mathrm{~m}$
Explanation:
B As we know, Time period of satellite is given by, $\mathrm{T}=\frac{2 \pi\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)}{\mathrm{v}}$ Orbital velocity of satellite is given by, $v=\sqrt{\frac{\mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}+\mathrm{h}}}$ Putting the value of $\mathrm{v}$ in equation (i), time period of satellite, $\mathrm{T}=\frac{2 \pi\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)}{\sqrt{\frac{\mathrm{GM}_{\mathrm{E}}}{\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)}}}$ $=\frac{2 \pi\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)^{\frac{3}{2}}}{\sqrt{\mathrm{GM}_{\mathrm{E}}}}$ Squaring both sides, we get - $\mathrm{T}^{2}=\frac{4 \pi^{2}\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)^{3}}{\mathrm{GM}_{\mathrm{E}}}$ $\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)^{3}=\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{T}^{2}}{4 \pi^{2}}$ $\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)=\left(\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{T}^{2}}{4 \pi^{2}}\right)^{1 / 3}$ $\mathrm{~h}=\left(\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{T}^{2}}{4 \pi^{2}}\right)^{1 / 3}-\mathrm{R}_{\mathrm{E}}$ Here, $\quad M_{E}=6 \times 10^{24} \mathrm{~kg}$ $\mathrm{R}_{\mathrm{e}}=6400 \mathrm{~km}=6400 \times 10^{3} \mathrm{~m}$ $=6.4 \times 10^{6} \mathrm{~m}$ $\mathrm{T}=24 \mathrm{~h}=24 \times 60 \times 60 \mathrm{~s}=86400 \mathrm{~s}$ $\mathrm{G}=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}$ On substituting the given values in equation (ii), we get $\mathrm{h}=\left(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times(86400)^{2}}{4 \times(3.14)^{2}}\right)^{1 / 3}-6.4 \times 10^{6}$ $\mathrm{~h}=4.21 \times 10^{7}-6.4 \times 10^{6}$ $=3.57 \times 10^{7} \mathrm{~m}$
AP EAMCET-24.04.2018
Gravitation
138647
A planet is revolving round the sun of mass ' $M$ ' in an elliptical orbit with semi-major axis ' $a$ '. The speed of the planer when it is a distance ' $r$ ' from the sun is. (G - Universal gravitational constant)
A In elliptical orbit, total energy. $\mathrm{E}=\frac{-\mathrm{GMm}}{2 \mathrm{a}} \quad(\mathrm{m}=\text { mass of planet })$ From conservation of mechanical energy $\text { K.E }+ \text { P.E. }=E$ $\frac{1}{2} m v^{2}-\frac{G M m}{r}=\frac{-G M m}{2 a}$ $\frac{1}{2} m v^{2}=\frac{G M m}{r}-\frac{G M m}{2 a}$ $\frac{1}{2} v^{2}=G M\left(\frac{1}{r}-\frac{1}{2 a}\right)$ $v=\sqrt{G M\left(\frac{2}{r}-\frac{1}{a}\right)}$
AP EAMCET (Medical)-24.04.2019
Gravitation
138648
The time period of moon around the earth is 28 days. If the mass of the earth is doubled, without any change in the distance of the moon from the earth. New time period of revolution of the moon is
1 $28 \sqrt{2}$ days
2 7 days
3 $14 \sqrt{2}$ days
4 14 days
Explanation:
C Given, $\mathrm{T}=28$ days We know, $\mathrm{T}^{2}=\frac{4 \pi^{2}}{\mathrm{GM}} \mathrm{r}^{3}$ When the mass of the earth doubled, $\mathrm{T}^{\prime 2}=\frac{4 \pi^{2}}{\mathrm{G} \cdot 2 \mathrm{M}} \mathrm{r}^{3}$ Dividing equation (ii) by (i), we get - $\therefore \quad \quad \frac{\mathrm{T}^{\prime 2}}{\mathrm{~T}^{2}} =\frac{\frac{4 \pi^{2}}{\mathrm{GM}} \cdot \frac{\mathrm{r}^{3}}{2}}{\frac{4 \pi^{2}}{\mathrm{GM}} \cdot \mathrm{r}^{3}}$ $\mathrm{~T}^{\prime 2} =\mathrm{T}^{2} / 2$ $\mathrm{~T}^{\prime} =\frac{\mathrm{T}}{\sqrt{2}}$ $\mathrm{~T}^{\prime} =\frac{28}{\sqrt{2}}$ $\mathrm{~T}^{\prime} =14 \sqrt{2}$
AP EAMCET-24.04.2019
Gravitation
138649
A number of planets are revolving around the Sun. Time period is ' $T$ ' and average orbital radius of a planet is ' $R$ '. A graph is drawn between $\log T$ on the $\mathrm{Y}$-axis and $\log \mathrm{R}$ on the $\mathrm{X}$-axis with the origin at $(0,0)$. The graph is a
1 Straight line with slope $\frac{3}{2}$ and passing through origin
2 Straight line with slope $\frac{3}{2}$ and not passing through origin
3 Parabola
4 Ellipse
Explanation:
B Using Kepler's $3^{\text {rd }}$ law, $T^{2} \propto R^{3}$ $T \propto R^{3 / 2}$ $T=K R^{3 / 2}$ Taking $\log$ of both side $\log \mathrm{T}=\log \left(\mathrm{KR}^{3 / 2}\right)$ $\log \mathrm{T}=\log \mathrm{K}+\frac{3}{2} \log \mathrm{R}$ $\log \mathrm{T}=\frac{3}{2} \log \mathrm{R}+\log \mathrm{K}$
138644
The distance between Sun and Earth is R. The duration of year if the distance between Sun and Earth becomes $3 R$ will be:
1 $\sqrt{3}$ years
2 3 years
3 8 years
4 $3 \sqrt{3}$ years
Explanation:
D Given, $\mathrm{T}=1$ year According to Kepler's third law $\mathrm{T}^{2} \propto \mathrm{r}^{3}$ If the distance between Sun and Earth is R $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ If the distance between Sun and Earth is $3 R$ $\left(\mathrm{T}^{\prime}\right)^{2} \propto(3 \mathrm{R})^{3}$ Dividing equation (ii) by (i), we get - $\left(\frac{\mathrm{T}^{\prime}}{\mathrm{T}}\right)^{2}=\left(\frac{3 \mathrm{R}}{\mathrm{R}}\right)^{3}$ $\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=(3)^{3 / 2}$ $\mathrm{~T}^{\prime}=3 \sqrt{3} \mathrm{~T}$ So, $T^{\prime}=3 \sqrt{3}$ years
JEE Main-24.06.2022
Gravitation
138646
A satellite is to be placed in equatorial geostationary orbit around the Earth for communication purpose. The height of such satellite is $\left(M_{E}=6 \times 10^{24} \mathrm{~kg}, R_{E}=6400 \mathrm{~km}\right)$
1 $3.57 \times 10^{8} \mathrm{~m}$
2 $3.57 \times 10^{7} \mathrm{~m}$
3 $3.57 \times 10^{5} \mathrm{~m}$
4 $3.57 \times 10^{6} \mathrm{~m}$
Explanation:
B As we know, Time period of satellite is given by, $\mathrm{T}=\frac{2 \pi\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)}{\mathrm{v}}$ Orbital velocity of satellite is given by, $v=\sqrt{\frac{\mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}+\mathrm{h}}}$ Putting the value of $\mathrm{v}$ in equation (i), time period of satellite, $\mathrm{T}=\frac{2 \pi\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)}{\sqrt{\frac{\mathrm{GM}_{\mathrm{E}}}{\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)}}}$ $=\frac{2 \pi\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)^{\frac{3}{2}}}{\sqrt{\mathrm{GM}_{\mathrm{E}}}}$ Squaring both sides, we get - $\mathrm{T}^{2}=\frac{4 \pi^{2}\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)^{3}}{\mathrm{GM}_{\mathrm{E}}}$ $\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)^{3}=\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{T}^{2}}{4 \pi^{2}}$ $\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)=\left(\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{T}^{2}}{4 \pi^{2}}\right)^{1 / 3}$ $\mathrm{~h}=\left(\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{T}^{2}}{4 \pi^{2}}\right)^{1 / 3}-\mathrm{R}_{\mathrm{E}}$ Here, $\quad M_{E}=6 \times 10^{24} \mathrm{~kg}$ $\mathrm{R}_{\mathrm{e}}=6400 \mathrm{~km}=6400 \times 10^{3} \mathrm{~m}$ $=6.4 \times 10^{6} \mathrm{~m}$ $\mathrm{T}=24 \mathrm{~h}=24 \times 60 \times 60 \mathrm{~s}=86400 \mathrm{~s}$ $\mathrm{G}=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}$ On substituting the given values in equation (ii), we get $\mathrm{h}=\left(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times(86400)^{2}}{4 \times(3.14)^{2}}\right)^{1 / 3}-6.4 \times 10^{6}$ $\mathrm{~h}=4.21 \times 10^{7}-6.4 \times 10^{6}$ $=3.57 \times 10^{7} \mathrm{~m}$
AP EAMCET-24.04.2018
Gravitation
138647
A planet is revolving round the sun of mass ' $M$ ' in an elliptical orbit with semi-major axis ' $a$ '. The speed of the planer when it is a distance ' $r$ ' from the sun is. (G - Universal gravitational constant)
A In elliptical orbit, total energy. $\mathrm{E}=\frac{-\mathrm{GMm}}{2 \mathrm{a}} \quad(\mathrm{m}=\text { mass of planet })$ From conservation of mechanical energy $\text { K.E }+ \text { P.E. }=E$ $\frac{1}{2} m v^{2}-\frac{G M m}{r}=\frac{-G M m}{2 a}$ $\frac{1}{2} m v^{2}=\frac{G M m}{r}-\frac{G M m}{2 a}$ $\frac{1}{2} v^{2}=G M\left(\frac{1}{r}-\frac{1}{2 a}\right)$ $v=\sqrt{G M\left(\frac{2}{r}-\frac{1}{a}\right)}$
AP EAMCET (Medical)-24.04.2019
Gravitation
138648
The time period of moon around the earth is 28 days. If the mass of the earth is doubled, without any change in the distance of the moon from the earth. New time period of revolution of the moon is
1 $28 \sqrt{2}$ days
2 7 days
3 $14 \sqrt{2}$ days
4 14 days
Explanation:
C Given, $\mathrm{T}=28$ days We know, $\mathrm{T}^{2}=\frac{4 \pi^{2}}{\mathrm{GM}} \mathrm{r}^{3}$ When the mass of the earth doubled, $\mathrm{T}^{\prime 2}=\frac{4 \pi^{2}}{\mathrm{G} \cdot 2 \mathrm{M}} \mathrm{r}^{3}$ Dividing equation (ii) by (i), we get - $\therefore \quad \quad \frac{\mathrm{T}^{\prime 2}}{\mathrm{~T}^{2}} =\frac{\frac{4 \pi^{2}}{\mathrm{GM}} \cdot \frac{\mathrm{r}^{3}}{2}}{\frac{4 \pi^{2}}{\mathrm{GM}} \cdot \mathrm{r}^{3}}$ $\mathrm{~T}^{\prime 2} =\mathrm{T}^{2} / 2$ $\mathrm{~T}^{\prime} =\frac{\mathrm{T}}{\sqrt{2}}$ $\mathrm{~T}^{\prime} =\frac{28}{\sqrt{2}}$ $\mathrm{~T}^{\prime} =14 \sqrt{2}$
AP EAMCET-24.04.2019
Gravitation
138649
A number of planets are revolving around the Sun. Time period is ' $T$ ' and average orbital radius of a planet is ' $R$ '. A graph is drawn between $\log T$ on the $\mathrm{Y}$-axis and $\log \mathrm{R}$ on the $\mathrm{X}$-axis with the origin at $(0,0)$. The graph is a
1 Straight line with slope $\frac{3}{2}$ and passing through origin
2 Straight line with slope $\frac{3}{2}$ and not passing through origin
3 Parabola
4 Ellipse
Explanation:
B Using Kepler's $3^{\text {rd }}$ law, $T^{2} \propto R^{3}$ $T \propto R^{3 / 2}$ $T=K R^{3 / 2}$ Taking $\log$ of both side $\log \mathrm{T}=\log \left(\mathrm{KR}^{3 / 2}\right)$ $\log \mathrm{T}=\log \mathrm{K}+\frac{3}{2} \log \mathrm{R}$ $\log \mathrm{T}=\frac{3}{2} \log \mathrm{R}+\log \mathrm{K}$
138644
The distance between Sun and Earth is R. The duration of year if the distance between Sun and Earth becomes $3 R$ will be:
1 $\sqrt{3}$ years
2 3 years
3 8 years
4 $3 \sqrt{3}$ years
Explanation:
D Given, $\mathrm{T}=1$ year According to Kepler's third law $\mathrm{T}^{2} \propto \mathrm{r}^{3}$ If the distance between Sun and Earth is R $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ If the distance between Sun and Earth is $3 R$ $\left(\mathrm{T}^{\prime}\right)^{2} \propto(3 \mathrm{R})^{3}$ Dividing equation (ii) by (i), we get - $\left(\frac{\mathrm{T}^{\prime}}{\mathrm{T}}\right)^{2}=\left(\frac{3 \mathrm{R}}{\mathrm{R}}\right)^{3}$ $\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=(3)^{3 / 2}$ $\mathrm{~T}^{\prime}=3 \sqrt{3} \mathrm{~T}$ So, $T^{\prime}=3 \sqrt{3}$ years
JEE Main-24.06.2022
Gravitation
138646
A satellite is to be placed in equatorial geostationary orbit around the Earth for communication purpose. The height of such satellite is $\left(M_{E}=6 \times 10^{24} \mathrm{~kg}, R_{E}=6400 \mathrm{~km}\right)$
1 $3.57 \times 10^{8} \mathrm{~m}$
2 $3.57 \times 10^{7} \mathrm{~m}$
3 $3.57 \times 10^{5} \mathrm{~m}$
4 $3.57 \times 10^{6} \mathrm{~m}$
Explanation:
B As we know, Time period of satellite is given by, $\mathrm{T}=\frac{2 \pi\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)}{\mathrm{v}}$ Orbital velocity of satellite is given by, $v=\sqrt{\frac{\mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}+\mathrm{h}}}$ Putting the value of $\mathrm{v}$ in equation (i), time period of satellite, $\mathrm{T}=\frac{2 \pi\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)}{\sqrt{\frac{\mathrm{GM}_{\mathrm{E}}}{\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)}}}$ $=\frac{2 \pi\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)^{\frac{3}{2}}}{\sqrt{\mathrm{GM}_{\mathrm{E}}}}$ Squaring both sides, we get - $\mathrm{T}^{2}=\frac{4 \pi^{2}\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)^{3}}{\mathrm{GM}_{\mathrm{E}}}$ $\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)^{3}=\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{T}^{2}}{4 \pi^{2}}$ $\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)=\left(\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{T}^{2}}{4 \pi^{2}}\right)^{1 / 3}$ $\mathrm{~h}=\left(\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{T}^{2}}{4 \pi^{2}}\right)^{1 / 3}-\mathrm{R}_{\mathrm{E}}$ Here, $\quad M_{E}=6 \times 10^{24} \mathrm{~kg}$ $\mathrm{R}_{\mathrm{e}}=6400 \mathrm{~km}=6400 \times 10^{3} \mathrm{~m}$ $=6.4 \times 10^{6} \mathrm{~m}$ $\mathrm{T}=24 \mathrm{~h}=24 \times 60 \times 60 \mathrm{~s}=86400 \mathrm{~s}$ $\mathrm{G}=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}$ On substituting the given values in equation (ii), we get $\mathrm{h}=\left(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times(86400)^{2}}{4 \times(3.14)^{2}}\right)^{1 / 3}-6.4 \times 10^{6}$ $\mathrm{~h}=4.21 \times 10^{7}-6.4 \times 10^{6}$ $=3.57 \times 10^{7} \mathrm{~m}$
AP EAMCET-24.04.2018
Gravitation
138647
A planet is revolving round the sun of mass ' $M$ ' in an elliptical orbit with semi-major axis ' $a$ '. The speed of the planer when it is a distance ' $r$ ' from the sun is. (G - Universal gravitational constant)
A In elliptical orbit, total energy. $\mathrm{E}=\frac{-\mathrm{GMm}}{2 \mathrm{a}} \quad(\mathrm{m}=\text { mass of planet })$ From conservation of mechanical energy $\text { K.E }+ \text { P.E. }=E$ $\frac{1}{2} m v^{2}-\frac{G M m}{r}=\frac{-G M m}{2 a}$ $\frac{1}{2} m v^{2}=\frac{G M m}{r}-\frac{G M m}{2 a}$ $\frac{1}{2} v^{2}=G M\left(\frac{1}{r}-\frac{1}{2 a}\right)$ $v=\sqrt{G M\left(\frac{2}{r}-\frac{1}{a}\right)}$
AP EAMCET (Medical)-24.04.2019
Gravitation
138648
The time period of moon around the earth is 28 days. If the mass of the earth is doubled, without any change in the distance of the moon from the earth. New time period of revolution of the moon is
1 $28 \sqrt{2}$ days
2 7 days
3 $14 \sqrt{2}$ days
4 14 days
Explanation:
C Given, $\mathrm{T}=28$ days We know, $\mathrm{T}^{2}=\frac{4 \pi^{2}}{\mathrm{GM}} \mathrm{r}^{3}$ When the mass of the earth doubled, $\mathrm{T}^{\prime 2}=\frac{4 \pi^{2}}{\mathrm{G} \cdot 2 \mathrm{M}} \mathrm{r}^{3}$ Dividing equation (ii) by (i), we get - $\therefore \quad \quad \frac{\mathrm{T}^{\prime 2}}{\mathrm{~T}^{2}} =\frac{\frac{4 \pi^{2}}{\mathrm{GM}} \cdot \frac{\mathrm{r}^{3}}{2}}{\frac{4 \pi^{2}}{\mathrm{GM}} \cdot \mathrm{r}^{3}}$ $\mathrm{~T}^{\prime 2} =\mathrm{T}^{2} / 2$ $\mathrm{~T}^{\prime} =\frac{\mathrm{T}}{\sqrt{2}}$ $\mathrm{~T}^{\prime} =\frac{28}{\sqrt{2}}$ $\mathrm{~T}^{\prime} =14 \sqrt{2}$
AP EAMCET-24.04.2019
Gravitation
138649
A number of planets are revolving around the Sun. Time period is ' $T$ ' and average orbital radius of a planet is ' $R$ '. A graph is drawn between $\log T$ on the $\mathrm{Y}$-axis and $\log \mathrm{R}$ on the $\mathrm{X}$-axis with the origin at $(0,0)$. The graph is a
1 Straight line with slope $\frac{3}{2}$ and passing through origin
2 Straight line with slope $\frac{3}{2}$ and not passing through origin
3 Parabola
4 Ellipse
Explanation:
B Using Kepler's $3^{\text {rd }}$ law, $T^{2} \propto R^{3}$ $T \propto R^{3 / 2}$ $T=K R^{3 / 2}$ Taking $\log$ of both side $\log \mathrm{T}=\log \left(\mathrm{KR}^{3 / 2}\right)$ $\log \mathrm{T}=\log \mathrm{K}+\frac{3}{2} \log \mathrm{R}$ $\log \mathrm{T}=\frac{3}{2} \log \mathrm{R}+\log \mathrm{K}$
