A satellite of mass \( \frac{M}{2} \) is revolving around Earth in a circular orbit at a height of \( \frac{R}{3} \) from the Earth's surface. The angular momentum of the satellite is \( M \sqrt{\frac{GM R}{x}} \). The value of \( x \) is:
Show Hint
Correct Answer: 3
Approach Solution - 1
Step 1: Orbital radius
Given: \[ r = R + \frac{R}{3} = \frac{4R}{3} \]
Step 2: Orbital velocity using gravitational and centripetal force balance
For circular motion, gravitational force provides the centripetal force: \[ \frac{GMm}{r^2} = \frac{mv^2}{r} \] Simplifying, we get: \[ v = \sqrt{\frac{GM}{r}} \]
Step 3: Angular momentum of the satellite
The angular momentum \( L \) of the satellite is: \[ L = mvr \]
Step 4: Substituting given values
Satellite mass: \[ m = \frac{M}{2} \] Orbital radius: \[ r = \frac{4R}{3} \] Orbital speed: \[ v = \sqrt{\frac{GM}{\frac{4R}{3}}} = \sqrt{\frac{3GM}{4R}} \]
Step 5: Substitute into the formula for angular momentum
\[ L = \frac{M}{2} \cdot \frac{4R}{3} \cdot \sqrt{\frac{3GM}{4R}} \] Simplify: \[ L = \frac{2MR}{3} \sqrt{\frac{3GM}{4R}} \]
Step 6: Simplifying the square root
\[ \sqrt{\frac{3GM}{4R}} = \sqrt{\frac{3}{4}} \sqrt{\frac{GM}{R}} = \frac{\sqrt{3}}{2} \sqrt{\frac{GM}{R}} \] Therefore, \[ L = \frac{2MR}{3} \times \frac{\sqrt{3}}{2} \sqrt{\frac{GM}{R}} \] \[ L = \frac{M\sqrt{3}}{3} \sqrt{GMR} \] \[ L = \frac{M}{\sqrt{3}} \sqrt{GMR} \]
Step 7: Alternative given form of angular momentum
It is given that: \[ L = M \sqrt{\frac{GMR}{x}} \]
Step 8: Equating both expressions
\[ \frac{M}{\sqrt{3}} \sqrt{GMR} = M \sqrt{\frac{GMR}{x}} \] Cancel \( M \) and \( \sqrt{GMR} \) (since both are non-zero): \[ \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{x}} \]
Step 9: Squaring both sides
\[ \frac{1}{3} = \frac{1}{x} \Rightarrow x = 3 \]
Final Answer:
\[ \boxed{x = 3} \]
Concept :
- Gravitational force acts as the centripetal force for orbital motion.
- Orbital velocity for a circular orbit: \( v = \sqrt{\frac{GM}{r}} \).
- Angular momentum of a satellite: \( L = mvr \).
- Simplify carefully when combining constants under square roots.
Approach Solution -2
The angular momentum \( L \) of a satellite in a circular orbit is given by:
\[ L = mvr, \]
Where:
- \( m \) is the mass of the satellite
- \( v \) is the orbital velocity
- \( r \) is the radius of the orbit (from the Earth's center)
Step 1: Orbital Velocity
The orbital velocity is given by:
\[ v = \sqrt{\frac{GM}{r}}, \]
Where \( G \) is the gravitational constant, and \( M \) is the Earth's mass.
Step 2: Orbit Radius
The satellite is at a height of \( \frac{R}{3} \) from the Earth's surface. So the total distance from the Earth's center is:
\[ r = R + \frac{R}{3} = \frac{4R}{3}. \]
Step 3: Substitute into Angular Momentum
Now substitute into the formula for \( L \):
\[ L = m \cdot v \cdot r = m \cdot \sqrt{\frac{GM}{r}} \cdot r. \]
Substitute \( r = \frac{4R}{3} \):
\[ L = m \cdot \sqrt{\frac{GM}{\frac{4R}{3}}} \cdot \frac{4R}{3} = m \cdot \sqrt{\frac{3GM}{4R}} \cdot \frac{4R}{3}. \]
Step 4: Use Given Satellite Mass
The satellite has mass \( m = \frac{M}{2} \), so we get:
\[ L = \frac{M}{2} \cdot \sqrt{\frac{3GM}{4R}} \cdot \frac{4R}{3}. \]
Step 5: Simplify
Simplify the constants:
\[ L = \frac{M}{2} \cdot \frac{4R}{3} \cdot \sqrt{\frac{3GM}{4R}}. \]
Combine terms under the square root and outside:
\[ L = M \cdot \sqrt{\frac{GM R}{3}}. \]
Step 6: Final Answer
The angular momentum is:
\[ L = M \cdot \sqrt{\frac{GMR}{3}}, \]
Therefore, the value of \( x = 3 \).
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Satellite Motion and Angular Momentum Questions
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.