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A body of mass 1 kg is moving under a central force in an elliptic orbit with semi major axis 1000 m and semi minor axis 100 m. The orbital angular momentum of the body is 100 kg m2 s-1. The time period of motion of the body is __________ hours.
(Specify your answer in hours upto two digits after the decimal point.)
(Specify your answer in hours upto two digits after the decimal point.)
Correct answer is between '1.70,1.80'. Can you explain this answer?
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A body of mass 1 kg is moving under a central force in an elliptic orb...
Given data:
- 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 m² s⁻¹
We need to find the time period of motion of the body.
1. Calculating Eccentricity (e):
The eccentricity of an elliptical orbit can be calculated using the formula:
e = √(1 - (b²/a²))
Given that a = 1000 m and b = 100 m, we can calculate the eccentricity as follows:
e = √(1 - (100²/1000²))
e = √(1 - 0.01)
e = √0.99
2. Calculating Angular Momentum (L):
The orbital angular momentum of a body moving in an elliptical orbit can be calculated using the formula:
L = m√(GMea(1 - e²))
Given that L = 100 kg m² s⁻¹, m = 1 kg, G = 6.67 × 10⁻¹¹ N m² kg⁻², and Me = 5.97 × 10²⁴ kg (mass of the Earth), we can rearrange the formula to solve for a:
L = m√(GMea(1 - e²))
100 = 1√(6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × a(1 - 0.99))
100 = √(4.013 × 10¹⁴ × a(1 - 0.99))
100 = √(4.013 × 10¹⁴ × a(0.01))
√(4.013 × 10¹⁴ × a) = 100/√0.01
4.013 × 10¹⁴ × a = (100/√0.01)²
4.013 × 10¹⁴ × a = 10⁶ × 10⁶
4.013 × 10¹⁴ × a = 10¹²
a = 10¹² / (4.013 × 10¹⁴)
a ≈ 2.49 × 10⁻³ m
3. Calculating the Time Period (T):
The time period of motion of a body in an elliptical orbit can be calculated using the formula:
T = 2π√(a³ / GMe)
Given that a = 2.49 × 10⁻³ m, G = 6.67 × 10⁻¹¹ N m² kg⁻², and Me = 5.97 × 10²⁴ kg, we can calculate the time period as follows:
T = 2π√(a³ / GMe)
T = 2π√((2.49 × 10⁻³)³ / (6.67 × 10⁻¹¹ × 5.97 × 10²⁴))
T = 2π√(15.5025 × 10⁻⁹ / 3.9822 × 10³⁵)
T = 2π
- 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 m² s⁻¹
We need to find the time period of motion of the body.
1. Calculating Eccentricity (e):
The eccentricity of an elliptical orbit can be calculated using the formula:
e = √(1 - (b²/a²))
Given that a = 1000 m and b = 100 m, we can calculate the eccentricity as follows:
e = √(1 - (100²/1000²))
e = √(1 - 0.01)
e = √0.99
2. Calculating Angular Momentum (L):
The orbital angular momentum of a body moving in an elliptical orbit can be calculated using the formula:
L = m√(GMea(1 - e²))
Given that L = 100 kg m² s⁻¹, m = 1 kg, G = 6.67 × 10⁻¹¹ N m² kg⁻², and Me = 5.97 × 10²⁴ kg (mass of the Earth), we can rearrange the formula to solve for a:
L = m√(GMea(1 - e²))
100 = 1√(6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × a(1 - 0.99))
100 = √(4.013 × 10¹⁴ × a(1 - 0.99))
100 = √(4.013 × 10¹⁴ × a(0.01))
√(4.013 × 10¹⁴ × a) = 100/√0.01
4.013 × 10¹⁴ × a = (100/√0.01)²
4.013 × 10¹⁴ × a = 10⁶ × 10⁶
4.013 × 10¹⁴ × a = 10¹²
a = 10¹² / (4.013 × 10¹⁴)
a ≈ 2.49 × 10⁻³ m
3. Calculating the Time Period (T):
The time period of motion of a body in an elliptical orbit can be calculated using the formula:
T = 2π√(a³ / GMe)
Given that a = 2.49 × 10⁻³ m, G = 6.67 × 10⁻¹¹ N m² kg⁻², and Me = 5.97 × 10²⁴ kg, we can calculate the time period as follows:
T = 2π√(a³ / GMe)
T = 2π√((2.49 × 10⁻³)³ / (6.67 × 10⁻¹¹ × 5.97 × 10²⁴))
T = 2π√(15.5025 × 10⁻⁹ / 3.9822 × 10³⁵)
T = 2π
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A body of mass 1 kg is moving under a central force in an elliptic orbit with semi major axis 1000 m and semi minor axis 100 m. The orbital angular momentum of the body is 100 kg m2 s-1. The time period of motion of the body is __________ hours.(Specify your answer in hours upto two digits after the decimal point.)Correct answer is between '1.70,1.80'. Can you explain this answer?
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A body of mass 1 kg is moving under a central force in an elliptic orbit with semi major axis 1000 m and semi minor axis 100 m. The orbital angular momentum of the body is 100 kg m2 s-1. The time period of motion of the body is __________ hours.(Specify your answer in hours upto two digits after the decimal point.)Correct answer is between '1.70,1.80'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A body of mass 1 kg is moving under a central force in an elliptic orbit with semi major axis 1000 m and semi minor axis 100 m. The orbital angular momentum of the body is 100 kg m2 s-1. The time period of motion of the body is __________ hours.(Specify your answer in hours upto two digits after the decimal point.)Correct answer is between '1.70,1.80'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body of mass 1 kg is moving under a central force in an elliptic orbit with semi major axis 1000 m and semi minor axis 100 m. The orbital angular momentum of the body is 100 kg m2 s-1. The time period of motion of the body is __________ hours.(Specify your answer in hours upto two digits after the decimal point.)Correct answer is between '1.70,1.80'. Can you explain this answer?.
A body of mass 1 kg is moving under a central force in an elliptic orbit with semi major axis 1000 m and semi minor axis 100 m. The orbital angular momentum of the body is 100 kg m2 s-1. The time period of motion of the body is __________ hours.(Specify your answer in hours upto two digits after the decimal point.)Correct answer is between '1.70,1.80'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A body of mass 1 kg is moving under a central force in an elliptic orbit with semi major axis 1000 m and semi minor axis 100 m. The orbital angular momentum of the body is 100 kg m2 s-1. The time period of motion of the body is __________ hours.(Specify your answer in hours upto two digits after the decimal point.)Correct answer is between '1.70,1.80'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body of mass 1 kg is moving under a central force in an elliptic orbit with semi major axis 1000 m and semi minor axis 100 m. The orbital angular momentum of the body is 100 kg m2 s-1. The time period of motion of the body is __________ hours.(Specify your answer in hours upto two digits after the decimal point.)Correct answer is between '1.70,1.80'. Can you explain this answer?.
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