359963
If the earth is at one-fourth of its present distance from the sun, the duration of the year would be
1 Half the present year
2 One-eigth the present year
3 One-fourth the present year
4 One -sixteenth the present year
Explanation:
Time period is related with radius as \(\begin{aligned}& T^{2} \alpha r^{3} \\& \left(\dfrac{T_{1}}{T_{2}}\right)^{2}=\left(\dfrac{r_{1}}{r_{2}}\right)^{3} \\& \left(\dfrac{T_{1}}{T_{2}}\right)^{2}=\left(\dfrac{r_{1}}{\left(r_{1} / 4\right)}\right)^{3}=4^{3} \\& \dfrac{T_{1}}{T_{2}}=4^{3 / 2}=8 \Rightarrow T_{2}=\dfrac{T_{1}}{8}\end{aligned}\)
PHXI08:GRAVITATION
359964
A light planet is revolving around a massive star in a circular orbit of radius \(R\) with a period of revolution \(T\). If the force of attraction between planet and star is proportional to \(R^{-3 / 2}\) then choose the correct option
1 \(T^{2} \propto R^{7 / 2}\)
2 \(T^{2} \propto R^{3}\)
3 \(T^{2} \propto R^{5 / 2}\)
4 \(T^{2} \propto R^{3 / 2}\)
Explanation:
Force of attraction between planet and star, to revolve the planet in a circular orbit is \(\therefore F=\dfrac{G M m}{R^{2}}=\dfrac{m v^{2}}{R}\) As per question \(F \propto R^{-3 / 2}\) So, \(\dfrac{m v^{2}}{R} \propto R^{-3 / 2} \Rightarrow m R \omega^{2} \propto R^{-3 / 2}\) \(\therefore \omega^{2} \propto R^{-3 / 2} \cdot R^{-1} \Rightarrow \omega^{2} \propto R^{-5 / 2}\) \(\left(\dfrac{1}{T}\right)^{2} \propto \dfrac{1}{R^{5 / 2}}\left(\because \omega=\dfrac{2 \pi}{T}\right)\) So, \(T^{2} \propto R^{5 / 2}\)
JEE - 2024
PHXI08:GRAVITATION
359965
A particle is moving with a uniform speed in a circular orbit of radius \(R\) in a central force inversely proportional to the \(n^{\text {th }}\) power of \(R\). If the period of rotation of the particle is \(T\), then,
359963
If the earth is at one-fourth of its present distance from the sun, the duration of the year would be
1 Half the present year
2 One-eigth the present year
3 One-fourth the present year
4 One -sixteenth the present year
Explanation:
Time period is related with radius as \(\begin{aligned}& T^{2} \alpha r^{3} \\& \left(\dfrac{T_{1}}{T_{2}}\right)^{2}=\left(\dfrac{r_{1}}{r_{2}}\right)^{3} \\& \left(\dfrac{T_{1}}{T_{2}}\right)^{2}=\left(\dfrac{r_{1}}{\left(r_{1} / 4\right)}\right)^{3}=4^{3} \\& \dfrac{T_{1}}{T_{2}}=4^{3 / 2}=8 \Rightarrow T_{2}=\dfrac{T_{1}}{8}\end{aligned}\)
PHXI08:GRAVITATION
359964
A light planet is revolving around a massive star in a circular orbit of radius \(R\) with a period of revolution \(T\). If the force of attraction between planet and star is proportional to \(R^{-3 / 2}\) then choose the correct option
1 \(T^{2} \propto R^{7 / 2}\)
2 \(T^{2} \propto R^{3}\)
3 \(T^{2} \propto R^{5 / 2}\)
4 \(T^{2} \propto R^{3 / 2}\)
Explanation:
Force of attraction between planet and star, to revolve the planet in a circular orbit is \(\therefore F=\dfrac{G M m}{R^{2}}=\dfrac{m v^{2}}{R}\) As per question \(F \propto R^{-3 / 2}\) So, \(\dfrac{m v^{2}}{R} \propto R^{-3 / 2} \Rightarrow m R \omega^{2} \propto R^{-3 / 2}\) \(\therefore \omega^{2} \propto R^{-3 / 2} \cdot R^{-1} \Rightarrow \omega^{2} \propto R^{-5 / 2}\) \(\left(\dfrac{1}{T}\right)^{2} \propto \dfrac{1}{R^{5 / 2}}\left(\because \omega=\dfrac{2 \pi}{T}\right)\) So, \(T^{2} \propto R^{5 / 2}\)
JEE - 2024
PHXI08:GRAVITATION
359965
A particle is moving with a uniform speed in a circular orbit of radius \(R\) in a central force inversely proportional to the \(n^{\text {th }}\) power of \(R\). If the period of rotation of the particle is \(T\), then,
359963
If the earth is at one-fourth of its present distance from the sun, the duration of the year would be
1 Half the present year
2 One-eigth the present year
3 One-fourth the present year
4 One -sixteenth the present year
Explanation:
Time period is related with radius as \(\begin{aligned}& T^{2} \alpha r^{3} \\& \left(\dfrac{T_{1}}{T_{2}}\right)^{2}=\left(\dfrac{r_{1}}{r_{2}}\right)^{3} \\& \left(\dfrac{T_{1}}{T_{2}}\right)^{2}=\left(\dfrac{r_{1}}{\left(r_{1} / 4\right)}\right)^{3}=4^{3} \\& \dfrac{T_{1}}{T_{2}}=4^{3 / 2}=8 \Rightarrow T_{2}=\dfrac{T_{1}}{8}\end{aligned}\)
PHXI08:GRAVITATION
359964
A light planet is revolving around a massive star in a circular orbit of radius \(R\) with a period of revolution \(T\). If the force of attraction between planet and star is proportional to \(R^{-3 / 2}\) then choose the correct option
1 \(T^{2} \propto R^{7 / 2}\)
2 \(T^{2} \propto R^{3}\)
3 \(T^{2} \propto R^{5 / 2}\)
4 \(T^{2} \propto R^{3 / 2}\)
Explanation:
Force of attraction between planet and star, to revolve the planet in a circular orbit is \(\therefore F=\dfrac{G M m}{R^{2}}=\dfrac{m v^{2}}{R}\) As per question \(F \propto R^{-3 / 2}\) So, \(\dfrac{m v^{2}}{R} \propto R^{-3 / 2} \Rightarrow m R \omega^{2} \propto R^{-3 / 2}\) \(\therefore \omega^{2} \propto R^{-3 / 2} \cdot R^{-1} \Rightarrow \omega^{2} \propto R^{-5 / 2}\) \(\left(\dfrac{1}{T}\right)^{2} \propto \dfrac{1}{R^{5 / 2}}\left(\because \omega=\dfrac{2 \pi}{T}\right)\) So, \(T^{2} \propto R^{5 / 2}\)
JEE - 2024
PHXI08:GRAVITATION
359965
A particle is moving with a uniform speed in a circular orbit of radius \(R\) in a central force inversely proportional to the \(n^{\text {th }}\) power of \(R\). If the period of rotation of the particle is \(T\), then,
359963
If the earth is at one-fourth of its present distance from the sun, the duration of the year would be
1 Half the present year
2 One-eigth the present year
3 One-fourth the present year
4 One -sixteenth the present year
Explanation:
Time period is related with radius as \(\begin{aligned}& T^{2} \alpha r^{3} \\& \left(\dfrac{T_{1}}{T_{2}}\right)^{2}=\left(\dfrac{r_{1}}{r_{2}}\right)^{3} \\& \left(\dfrac{T_{1}}{T_{2}}\right)^{2}=\left(\dfrac{r_{1}}{\left(r_{1} / 4\right)}\right)^{3}=4^{3} \\& \dfrac{T_{1}}{T_{2}}=4^{3 / 2}=8 \Rightarrow T_{2}=\dfrac{T_{1}}{8}\end{aligned}\)
PHXI08:GRAVITATION
359964
A light planet is revolving around a massive star in a circular orbit of radius \(R\) with a period of revolution \(T\). If the force of attraction between planet and star is proportional to \(R^{-3 / 2}\) then choose the correct option
1 \(T^{2} \propto R^{7 / 2}\)
2 \(T^{2} \propto R^{3}\)
3 \(T^{2} \propto R^{5 / 2}\)
4 \(T^{2} \propto R^{3 / 2}\)
Explanation:
Force of attraction between planet and star, to revolve the planet in a circular orbit is \(\therefore F=\dfrac{G M m}{R^{2}}=\dfrac{m v^{2}}{R}\) As per question \(F \propto R^{-3 / 2}\) So, \(\dfrac{m v^{2}}{R} \propto R^{-3 / 2} \Rightarrow m R \omega^{2} \propto R^{-3 / 2}\) \(\therefore \omega^{2} \propto R^{-3 / 2} \cdot R^{-1} \Rightarrow \omega^{2} \propto R^{-5 / 2}\) \(\left(\dfrac{1}{T}\right)^{2} \propto \dfrac{1}{R^{5 / 2}}\left(\because \omega=\dfrac{2 \pi}{T}\right)\) So, \(T^{2} \propto R^{5 / 2}\)
JEE - 2024
PHXI08:GRAVITATION
359965
A particle is moving with a uniform speed in a circular orbit of radius \(R\) in a central force inversely proportional to the \(n^{\text {th }}\) power of \(R\). If the period of rotation of the particle is \(T\), then,
