359805
Assertion : Two different planets can have same escape velocity Reason : Value of escape velocity is a universal constant.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
As, escape velocity \(=\sqrt{\dfrac{G M}{R}}\), so its value depends on mass of planet and radius of the planet. The two different planets have same escape velocity, when mass and radius are equal for both planets. So option (3) is correct
PHXI08:GRAVITATION
359806
A body is projected vertically from the surface of the earth of radius \('R'\) with velocity equal to half of the escape velocity. The maximum height reached by the body is
1 \(\dfrac{R}{5}\)
2 \(\dfrac{R}{3}\)
3 \(\dfrac{R}{2}\)
4 \(\dfrac{R}{4}\)
Explanation:
According to the law of conservation of energy \(\begin{aligned}& (K E+P E)_{\text {surface }}=(K E+P E)_{(\max \text { height })} \\& \Rightarrow \dfrac{1}{2} m v^{2}+\left(-\dfrac{G M m}{R}\right)=0+\left(-\dfrac{G M m}{R+h}\right)\end{aligned}\) Given, \(\quad v=\dfrac{1}{2} v_{e}=\dfrac{1}{2} \sqrt{\dfrac{2 G M}{R}}\) \(\begin{aligned}& \Rightarrow \dfrac{1}{2} m\left[\dfrac{1}{4} \cdot \dfrac{2 G M}{R}\right]-\dfrac{G M m}{R}=-\dfrac{G M m}{R+h} \\& \Rightarrow \dfrac{1}{4} \dfrac{G M m}{R}-\dfrac{G M m}{R}=-\dfrac{G M m}{R+h} \\& \Rightarrow \dfrac{3}{4 R}=\dfrac{1}{R+h} \\& \Rightarrow h=\dfrac{R}{3}\end{aligned}\)
PHXI08:GRAVITATION
359807
A space station is at a height equal to the radius of the Earth. If \(v_{E}\) is the escape velocity on the surface of the Earth, the same on the space station is_____times \(v_{E}\)
1 \(\dfrac{1}{2}\)
2 \(\dfrac{1}{4}\)
3 \(\dfrac{1}{\sqrt{2}}\)
4 \(\dfrac{1}{\sqrt{3}}\)
Explanation:
As the space station is at a height \(h\) from the surface of the earth, its distance from the centre of the Earth is \((R + h)\) Thus, the escape velocity on the space station is \(\begin{aligned}& v_{s}=\sqrt{\dfrac{2 G M}{(R+h)}}=\sqrt{\dfrac{2 G M}{(R+R)}}=\sqrt{\dfrac{2 G M}{2 R}}(\text { as } h=R) \\= & \dfrac{1}{\sqrt{2}}\left(\sqrt{\dfrac{2 G M}{R}}\right)=\dfrac{v_{E}}{\sqrt{2}}\end{aligned}\)
KCET - 2018
PHXI08:GRAVITATION
359808
The escape velocity from the Earth's surface is \(v\). The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is :
1 \(2 v\)
2 \(3 v\)
3 \(4 v\)
4 \(v\)
Explanation:
\(v_{e}=\sqrt{\dfrac{2 G M}{R}}\) \(=\sqrt{\dfrac{2 G \dfrac{4}{3} \pi R^{3} \rho}{R}}\) \(\begin{gathered}=R \sqrt{\dfrac{8}{3} \pi G \rho} \\v_{e} \propto R \\\dfrac{v}{v_{P}}=\dfrac{R}{4 R} \\v_{P}=4 v\end{gathered}\)
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PHXI08:GRAVITATION
359805
Assertion : Two different planets can have same escape velocity Reason : Value of escape velocity is a universal constant.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
As, escape velocity \(=\sqrt{\dfrac{G M}{R}}\), so its value depends on mass of planet and radius of the planet. The two different planets have same escape velocity, when mass and radius are equal for both planets. So option (3) is correct
PHXI08:GRAVITATION
359806
A body is projected vertically from the surface of the earth of radius \('R'\) with velocity equal to half of the escape velocity. The maximum height reached by the body is
1 \(\dfrac{R}{5}\)
2 \(\dfrac{R}{3}\)
3 \(\dfrac{R}{2}\)
4 \(\dfrac{R}{4}\)
Explanation:
According to the law of conservation of energy \(\begin{aligned}& (K E+P E)_{\text {surface }}=(K E+P E)_{(\max \text { height })} \\& \Rightarrow \dfrac{1}{2} m v^{2}+\left(-\dfrac{G M m}{R}\right)=0+\left(-\dfrac{G M m}{R+h}\right)\end{aligned}\) Given, \(\quad v=\dfrac{1}{2} v_{e}=\dfrac{1}{2} \sqrt{\dfrac{2 G M}{R}}\) \(\begin{aligned}& \Rightarrow \dfrac{1}{2} m\left[\dfrac{1}{4} \cdot \dfrac{2 G M}{R}\right]-\dfrac{G M m}{R}=-\dfrac{G M m}{R+h} \\& \Rightarrow \dfrac{1}{4} \dfrac{G M m}{R}-\dfrac{G M m}{R}=-\dfrac{G M m}{R+h} \\& \Rightarrow \dfrac{3}{4 R}=\dfrac{1}{R+h} \\& \Rightarrow h=\dfrac{R}{3}\end{aligned}\)
PHXI08:GRAVITATION
359807
A space station is at a height equal to the radius of the Earth. If \(v_{E}\) is the escape velocity on the surface of the Earth, the same on the space station is_____times \(v_{E}\)
1 \(\dfrac{1}{2}\)
2 \(\dfrac{1}{4}\)
3 \(\dfrac{1}{\sqrt{2}}\)
4 \(\dfrac{1}{\sqrt{3}}\)
Explanation:
As the space station is at a height \(h\) from the surface of the earth, its distance from the centre of the Earth is \((R + h)\) Thus, the escape velocity on the space station is \(\begin{aligned}& v_{s}=\sqrt{\dfrac{2 G M}{(R+h)}}=\sqrt{\dfrac{2 G M}{(R+R)}}=\sqrt{\dfrac{2 G M}{2 R}}(\text { as } h=R) \\= & \dfrac{1}{\sqrt{2}}\left(\sqrt{\dfrac{2 G M}{R}}\right)=\dfrac{v_{E}}{\sqrt{2}}\end{aligned}\)
KCET - 2018
PHXI08:GRAVITATION
359808
The escape velocity from the Earth's surface is \(v\). The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is :
1 \(2 v\)
2 \(3 v\)
3 \(4 v\)
4 \(v\)
Explanation:
\(v_{e}=\sqrt{\dfrac{2 G M}{R}}\) \(=\sqrt{\dfrac{2 G \dfrac{4}{3} \pi R^{3} \rho}{R}}\) \(\begin{gathered}=R \sqrt{\dfrac{8}{3} \pi G \rho} \\v_{e} \propto R \\\dfrac{v}{v_{P}}=\dfrac{R}{4 R} \\v_{P}=4 v\end{gathered}\)
359805
Assertion : Two different planets can have same escape velocity Reason : Value of escape velocity is a universal constant.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
As, escape velocity \(=\sqrt{\dfrac{G M}{R}}\), so its value depends on mass of planet and radius of the planet. The two different planets have same escape velocity, when mass and radius are equal for both planets. So option (3) is correct
PHXI08:GRAVITATION
359806
A body is projected vertically from the surface of the earth of radius \('R'\) with velocity equal to half of the escape velocity. The maximum height reached by the body is
1 \(\dfrac{R}{5}\)
2 \(\dfrac{R}{3}\)
3 \(\dfrac{R}{2}\)
4 \(\dfrac{R}{4}\)
Explanation:
According to the law of conservation of energy \(\begin{aligned}& (K E+P E)_{\text {surface }}=(K E+P E)_{(\max \text { height })} \\& \Rightarrow \dfrac{1}{2} m v^{2}+\left(-\dfrac{G M m}{R}\right)=0+\left(-\dfrac{G M m}{R+h}\right)\end{aligned}\) Given, \(\quad v=\dfrac{1}{2} v_{e}=\dfrac{1}{2} \sqrt{\dfrac{2 G M}{R}}\) \(\begin{aligned}& \Rightarrow \dfrac{1}{2} m\left[\dfrac{1}{4} \cdot \dfrac{2 G M}{R}\right]-\dfrac{G M m}{R}=-\dfrac{G M m}{R+h} \\& \Rightarrow \dfrac{1}{4} \dfrac{G M m}{R}-\dfrac{G M m}{R}=-\dfrac{G M m}{R+h} \\& \Rightarrow \dfrac{3}{4 R}=\dfrac{1}{R+h} \\& \Rightarrow h=\dfrac{R}{3}\end{aligned}\)
PHXI08:GRAVITATION
359807
A space station is at a height equal to the radius of the Earth. If \(v_{E}\) is the escape velocity on the surface of the Earth, the same on the space station is_____times \(v_{E}\)
1 \(\dfrac{1}{2}\)
2 \(\dfrac{1}{4}\)
3 \(\dfrac{1}{\sqrt{2}}\)
4 \(\dfrac{1}{\sqrt{3}}\)
Explanation:
As the space station is at a height \(h\) from the surface of the earth, its distance from the centre of the Earth is \((R + h)\) Thus, the escape velocity on the space station is \(\begin{aligned}& v_{s}=\sqrt{\dfrac{2 G M}{(R+h)}}=\sqrt{\dfrac{2 G M}{(R+R)}}=\sqrt{\dfrac{2 G M}{2 R}}(\text { as } h=R) \\= & \dfrac{1}{\sqrt{2}}\left(\sqrt{\dfrac{2 G M}{R}}\right)=\dfrac{v_{E}}{\sqrt{2}}\end{aligned}\)
KCET - 2018
PHXI08:GRAVITATION
359808
The escape velocity from the Earth's surface is \(v\). The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is :
1 \(2 v\)
2 \(3 v\)
3 \(4 v\)
4 \(v\)
Explanation:
\(v_{e}=\sqrt{\dfrac{2 G M}{R}}\) \(=\sqrt{\dfrac{2 G \dfrac{4}{3} \pi R^{3} \rho}{R}}\) \(\begin{gathered}=R \sqrt{\dfrac{8}{3} \pi G \rho} \\v_{e} \propto R \\\dfrac{v}{v_{P}}=\dfrac{R}{4 R} \\v_{P}=4 v\end{gathered}\)
359805
Assertion : Two different planets can have same escape velocity Reason : Value of escape velocity is a universal constant.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
As, escape velocity \(=\sqrt{\dfrac{G M}{R}}\), so its value depends on mass of planet and radius of the planet. The two different planets have same escape velocity, when mass and radius are equal for both planets. So option (3) is correct
PHXI08:GRAVITATION
359806
A body is projected vertically from the surface of the earth of radius \('R'\) with velocity equal to half of the escape velocity. The maximum height reached by the body is
1 \(\dfrac{R}{5}\)
2 \(\dfrac{R}{3}\)
3 \(\dfrac{R}{2}\)
4 \(\dfrac{R}{4}\)
Explanation:
According to the law of conservation of energy \(\begin{aligned}& (K E+P E)_{\text {surface }}=(K E+P E)_{(\max \text { height })} \\& \Rightarrow \dfrac{1}{2} m v^{2}+\left(-\dfrac{G M m}{R}\right)=0+\left(-\dfrac{G M m}{R+h}\right)\end{aligned}\) Given, \(\quad v=\dfrac{1}{2} v_{e}=\dfrac{1}{2} \sqrt{\dfrac{2 G M}{R}}\) \(\begin{aligned}& \Rightarrow \dfrac{1}{2} m\left[\dfrac{1}{4} \cdot \dfrac{2 G M}{R}\right]-\dfrac{G M m}{R}=-\dfrac{G M m}{R+h} \\& \Rightarrow \dfrac{1}{4} \dfrac{G M m}{R}-\dfrac{G M m}{R}=-\dfrac{G M m}{R+h} \\& \Rightarrow \dfrac{3}{4 R}=\dfrac{1}{R+h} \\& \Rightarrow h=\dfrac{R}{3}\end{aligned}\)
PHXI08:GRAVITATION
359807
A space station is at a height equal to the radius of the Earth. If \(v_{E}\) is the escape velocity on the surface of the Earth, the same on the space station is_____times \(v_{E}\)
1 \(\dfrac{1}{2}\)
2 \(\dfrac{1}{4}\)
3 \(\dfrac{1}{\sqrt{2}}\)
4 \(\dfrac{1}{\sqrt{3}}\)
Explanation:
As the space station is at a height \(h\) from the surface of the earth, its distance from the centre of the Earth is \((R + h)\) Thus, the escape velocity on the space station is \(\begin{aligned}& v_{s}=\sqrt{\dfrac{2 G M}{(R+h)}}=\sqrt{\dfrac{2 G M}{(R+R)}}=\sqrt{\dfrac{2 G M}{2 R}}(\text { as } h=R) \\= & \dfrac{1}{\sqrt{2}}\left(\sqrt{\dfrac{2 G M}{R}}\right)=\dfrac{v_{E}}{\sqrt{2}}\end{aligned}\)
KCET - 2018
PHXI08:GRAVITATION
359808
The escape velocity from the Earth's surface is \(v\). The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is :
1 \(2 v\)
2 \(3 v\)
3 \(4 v\)
4 \(v\)
Explanation:
\(v_{e}=\sqrt{\dfrac{2 G M}{R}}\) \(=\sqrt{\dfrac{2 G \dfrac{4}{3} \pi R^{3} \rho}{R}}\) \(\begin{gathered}=R \sqrt{\dfrac{8}{3} \pi G \rho} \\v_{e} \propto R \\\dfrac{v}{v_{P}}=\dfrac{R}{4 R} \\v_{P}=4 v\end{gathered}\)