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Body of mass 1 kg is moving under a central force in an elliptical orbit with semi-major axis 1000 m and semi-minor axis 100 m. The orbital angular momentum of the body is 100 kg m²s-¹. If time period of motion of the body is N times 100 π second, then find the value of N.
Double-digit integer (-99 to 99)
Type your answer here?
Double-digit integer (-99 to 99)
Type your answer here?
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Body of mass 1 kg is moving under a central force in an elliptical orb...
Introduction
To find the value of N, we need to determine the time period of the motion of a body in an elliptical orbit. Given the parameters of the orbit and the angular momentum, we can derive the necessary values.
Key Parameters
- Mass of the body (m) = 1 kg
- Semi-major axis (a) = 1000 m
- Semi-minor axis (b) = 100 m
- Orbital angular momentum (L) = 100 kg m²s-¹
Time Period of Motion
The time period (T) of an elliptical orbit can be calculated using Kepler's Third Law, which states:
T = 2π * √(a³ / (GM))
Where:
- G is the gravitational constant (6.674 × 10^-11 N m²/kg²)
- M is the mass of the central body.
However, we can also derive T using the properties of angular momentum:
Relating Angular Momentum to Time Period
The angular momentum L is given by:
L = m * v * r
For an elliptical orbit, we can also express it as:
L = m * (2πa / T) * a
From this, we can rearrange to find T:
T = 2π * (a² / L)
Substituting the known values:
- a = 1000 m
- L = 100 kg m²/s
Calculation of T
T = 2π * (1000² / 100)
T = 2π * 10000
T = 20000π seconds
Using the problem's context, we have T = N * 100π seconds.
Finding N
Now, equating:
20000π = N * 100π
By simplifying, we find:
N = 200
Conclusion
The value of N is 200, which is a double-digit integer.
To find the value of N, we need to determine the time period of the motion of a body in an elliptical orbit. Given the parameters of the orbit and the angular momentum, we can derive the necessary values.
Key Parameters
- Mass of the body (m) = 1 kg
- Semi-major axis (a) = 1000 m
- Semi-minor axis (b) = 100 m
- Orbital angular momentum (L) = 100 kg m²s-¹
Time Period of Motion
The time period (T) of an elliptical orbit can be calculated using Kepler's Third Law, which states:
T = 2π * √(a³ / (GM))
Where:
- G is the gravitational constant (6.674 × 10^-11 N m²/kg²)
- M is the mass of the central body.
However, we can also derive T using the properties of angular momentum:
Relating Angular Momentum to Time Period
The angular momentum L is given by:
L = m * v * r
For an elliptical orbit, we can also express it as:
L = m * (2πa / T) * a
From this, we can rearrange to find T:
T = 2π * (a² / L)
Substituting the known values:
- a = 1000 m
- L = 100 kg m²/s
Calculation of T
T = 2π * (1000² / 100)
T = 2π * 10000
T = 20000π seconds
Using the problem's context, we have T = N * 100π seconds.
Finding N
Now, equating:
20000π = N * 100π
By simplifying, we find:
N = 200
Conclusion
The value of N is 200, which is a double-digit integer.
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Body of mass 1 kg is moving under a central force in an elliptical orbit with semi-major axis 1000 m and semi-minor axis 100 m. The orbital angular momentum of the body is 100 kg m²s-¹. If time period of motion of the body is N times 100 π second, then find the value of N.Double-digit integer (-99 to 99)Type your answer here?
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Body of mass 1 kg is moving under a central force in an elliptical orbit with semi-major axis 1000 m and semi-minor axis 100 m. The orbital angular momentum of the body is 100 kg m²s-¹. If time period of motion of the body is N times 100 π second, then find the value of N.Double-digit integer (-99 to 99)Type your answer here? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about Body of mass 1 kg is moving under a central force in an elliptical orbit with semi-major axis 1000 m and semi-minor axis 100 m. The orbital angular momentum of the body is 100 kg m²s-¹. If time period of motion of the body is N times 100 π second, then find the value of N.Double-digit integer (-99 to 99)Type your answer here? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Body of mass 1 kg is moving under a central force in an elliptical orbit with semi-major axis 1000 m and semi-minor axis 100 m. The orbital angular momentum of the body is 100 kg m²s-¹. If time period of motion of the body is N times 100 π second, then find the value of N.Double-digit integer (-99 to 99)Type your answer here?.
Body of mass 1 kg is moving under a central force in an elliptical orbit with semi-major axis 1000 m and semi-minor axis 100 m. The orbital angular momentum of the body is 100 kg m²s-¹. If time period of motion of the body is N times 100 π second, then find the value of N.Double-digit integer (-99 to 99)Type your answer here? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about Body of mass 1 kg is moving under a central force in an elliptical orbit with semi-major axis 1000 m and semi-minor axis 100 m. The orbital angular momentum of the body is 100 kg m²s-¹. If time period of motion of the body is N times 100 π second, then find the value of N.Double-digit integer (-99 to 99)Type your answer here? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Body of mass 1 kg is moving under a central force in an elliptical orbit with semi-major axis 1000 m and semi-minor axis 100 m. The orbital angular momentum of the body is 100 kg m²s-¹. If time period of motion of the body is N times 100 π second, then find the value of N.Double-digit integer (-99 to 99)Type your answer here?.
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