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If due to air drag, the orbital radius of the earth decreases from \(R\) to \(R-\Delta R, \Delta R \ll R\), then the expression for increase in the orbital velocity \(\Delta v\) is (1) \(\frac{\Delta R}{2} \sqrt{\frac{G M}{R^{3}}}\) (2) \(-\frac{\Delta R}{2} \sqrt{\frac{G M}{R^{3}}}\) (3) \(\Delta R \sqrt{\frac{G M}{R^{3}}}\) (4) \(-\Delta R \sqrt{\frac{G M}{R^{3}}}\)

Short Answer

Expert verified
The correct answer is option (1): \( \frac{\Delta R}{2} \sqrt{\frac{G M}{R^{3}}} \).

Step by step solution

01

Understanding Initial Conditions

Initially, the earth orbits the sun in a circular orbit at radius \(R\) with velocity \(v\). The gravitational force supplies the necessary centripetal force such that \(\frac{G M m}{R^2} = \frac{m v^2}{R}\), where \(G\) is the gravitational constant, \(M\) is the mass of the Sun, and \(m\) is the mass of the Earth.
02

Calculate Initial Orbital Velocity

From the balance of forces, we find the initial orbital velocity \(v\) as \(v = \sqrt{\frac{GM}{R}}\).
03

Understanding Change in Conditions

The decrease in radius due to air drag means a new orbital radius of \(R - \Delta R\). The new velocity is \(v + \Delta v\). The gravitational force at this new radius must provide the required centripetal force for this new orbit.
04

Calculate New Orbital Velocity

The new gravitational force is \(\frac{GMm}{(R - \Delta R)^2}\), and the new centripetal force is \(\frac{m(v + \Delta v)^2}{R - \Delta R}\). Equating the two and simplifying using \(\Delta R \ll R\), we assume \( (R - \Delta R)^2 \approx R^2 - 2R \Delta R\):\[\frac{GM}{R^2 - 2R \Delta R} \approx \frac{(v + \Delta v)^2}{R - \Delta R}\].
05

Approximate Changes and Simplify

Assuming \(\Delta v \ll v\) and \(\Delta R \ll R\), expand \((R - \Delta R)^2 \approx R^2 - 2R \Delta R\), and using a first-order expansion \( (v + \Delta v)^2 \approx v^2 + 2v\Delta v\).Substituting and solving: \[\frac{GM}{R^2 - 2R \Delta R} \approx \frac{v^2 + 2v \Delta v}{R}\].
06

Solve for Change in Velocity

Use the identity \( v^2 = \frac{GM}{R} \), and simplify for small \(\Delta R\):\[2v \Delta v \approx \frac{GM}{R} \left(\frac{2R \Delta R}{R^2}\right)\].Thus, \(2v \Delta v = \frac{2GM \Delta R}{R^3}\), or \( \Delta v = \frac{GM \Delta R}{R^3 v}\).Substitute \(v = \sqrt{\frac{GM}{R}}\) resulting in:\[\Delta v = \frac{\Delta R}{2} \sqrt{\frac{GM}{R^3}}\].
07

Identify Correct Option

The derived expression \(\frac{\Delta R}{2} \sqrt{\frac{G M}{R^3}}\) matches option (1), thus the correct answer is (1).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Orbital Velocity
Orbital velocity is the speed at which a planet or any object orbits around a celestial body, such as the Earth around the Sun. It is determined by the balance of gravitational force pulling the object towards the center of the celestial body and the object’s inertia trying to move it away. This balance between forces ensures that the object remains in a stable orbit.
For any object in a circular orbit of radius \( R \), the orbital velocity \( v \) can be calculated using the formula \( v = \sqrt{\frac{GM}{R}} \), where \( G \) is the gravitational constant and \( M \) is the mass of the body being orbited, like the Sun. This equation arises from setting the gravitational force equal to the centripetal force needed for circular motion.
  • Orbital velocity varies based on the distance from the central body. Closer distances mean higher speeds are required to prevent falling into the central body.
  • The velocity ensures that the object stays in motion along a circular path rather than falling straight into the gravitational source or flying off into space.
Understanding orbital velocity helps us determine the conditions needed to keep satellites in orbit and calculate changes if our position relative to a celestial body changes.
Gravitational Force
Gravitational force is the natural phenomenon by which all objects with mass attract each other. For orbital mechanics, the gravitational force is critical as it acts as the central force keeping planets, moons, and artificial satellites in their respective orbits. This invisible force can be understood through Newton’s law of universal gravitation, which states that the force between two masses is proportional to their product and inversely proportional to the square of the distance between them.
The formula for gravitational force \( F_g \) is given by:\[F_g = \frac{G M m}{R^2} \]where:\
  • \( G \) is the gravitational constant,
  • \( M \) and \( m \) are the masses involved,
  • \( R \) is the distance between the masses' centers.
The gravitational force is what governs planetary motion and dictates the specific speed needed for an object to remain in a stable orbit, as seen in the orbital velocity concept.
Centripetal Force
Centripetal force is the force required to keep an object moving in a circular path, directed towards the center around which the object is moving. This force is crucial in orbital mechanics as it keeps celestial bodies and artificial satellites on their designated paths. The gravitational force acting on a planet or satellite in orbit serves as the centripetal force, ensuring continuous motion around the celestial body.
In the context of the Earth orbiting the Sun, the centripetal force needed is supplied entirely by the Sun’s gravitational pull. By balancing the gravitational force \( \frac{G M m}{R^2} \) against the necessary centripetal force \( \frac{m v^2}{R} \), we maintain stable orbits. This balance can be expressed as:\[\frac{G M m}{R^2} = \frac{m v^2}{R}\]This equation shows how gravity provides the centripetal force necessary for circular orbital motion.
  • Centripetal force is not a separate force but is the result of other forces, such as gravity, acting to maintain circular motion.
  • It is essential for understanding why orbits are circular and how objects maintain their paths around a greater mass.
Air Drag Effect
The air drag effect refers to the resistance an object encounters as it moves through an atmosphere. In the context of orbital mechanics, while space is largely a vacuum, objects in low Earth orbit still encounter thin atmospheric particles causing drag. This drag results in gradual energy loss, leading the object to lose altitude and alter its orbital path.
When an orbiting body, like a satellite, experiences air drag, its speed and altitude are affected. As air drag increases, the orbiting object slows down slightly and sinks to a lower orbit, where again it balances the gravitational and centripetal forces at a new, lower altitude. This causes changes in the orbital velocity, as seen when the radius changes from \( R \) to \( R - \Delta R \).
  • The drag effect is minimal in higher orbits where the atmosphere is very thin, but it's significant in lower orbits like those of many satellites.
  • Managing drag is vital for maintaining the longevity and proper functioning of satellites, requiring occasional adjustments to correct their orbits.
Understanding air drag helps in predicting changes over time in orbital paths due to atmospheric resistance.

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Most popular questions from this chapter

If \(g\) is the acceleration due to gravity on the earth's surface. the change in the potential energy of an object of mass \(m\) raised from the surface of the earth to a height equal to the radius \(R\) of the earth is (1) \(\frac{m g R}{2}\) (2) \(2 \mathrm{mg} R\) (3) \(m g R\) (4) \(-m g R\)

If both the mass and radius of the earth decrease by \(1 \%\), the value of (1) acceleration due to gravity would decrease by nearly \(1 \%\) (2) acceleration due to gravity would increase by \(1 \%\) (3) escape velocity from the earth's surface would decrease by \(1 \%\) (4) the gravitational potential energy of a body on earth's surface will remain unchanged

A skylab of mass \(m \mathrm{~kg}\) is first launched from the surface of the earth in a circular orbit of radius \(2 R\) (from the centre of the earth) and then it is shifted from this circular orbit to another circular orbit of radius \(3 R\). The minimum energy required to place the lab in the first orbit and to shift the lab from first orbit to the second orbit are (1) \(\frac{3}{4} m g R, \frac{m g R}{6}\) (2) \(\frac{3}{4} m g R, \frac{m g R}{12}\) (3) \(m g R, m g R\) (4) \(2 m g R, m g R\)

Two spherical bodies of masses \(M\) and \(5 M\) and radii \(R\) and \(2 R\), respectively, are released in free space with initial separation between their centres equal to \(12 R\). If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is (1) \(2.5 R\) (2) \(4.5 R\) (3) \(7.5 R\) (4) \(1.5 R\)

Two particles of mass \(m\) and \(4 m\) are at rest at an infinite separation. They move towards each other under mutual gravitational attraction. If \(G\) is the universal gravitational constant. Then at separation \(r\), (Assume zero reference potential energy at infinite separation) (1) the total energy of the two object is zero (2) their relative velocity of approach is \(\left(\frac{10 G m}{r}\right)^{1 / 2}\) in magnitude (3) the total kinetic energy of the object is \(\frac{4 G m^{2}}{r}\) (4) Net angular momentum of both the particles is zero about any point

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