According to Kepler's law of periods, the ratio of the orbital periods of two planets is related to the ratio of their semi-major axes. Specifically, the ratio of the periods (T₂/T₁) equals the ratio of their radii raised to the power of three-halves. In this case, if the radius of the second planetRead more
According to Kepler’s law of periods, the ratio of the orbital periods of two planets is related to the ratio of their semi-major axes. Specifically, the ratio of the periods (T₂/T₁) equals the ratio of their radii raised to the power of three-halves. In this case, if the radius of the second planet is four times that of the first, the calculation shows that T₂/T₁ equals eight.
Therefore, if the orbital period of the first planet is 1 day, the orbital period of the second planet would be 8 days, demonstrating the significant impact of radius on orbital time.
According to Kepler’s law of periods,
T₂/T₁ = (r₂/r₁)³/² = (4/1)³/² = 8
T₂ = 8T₁= 8 x 1 day = 8 days
A satellite orbiting Earth with a specific orbital radius and time period exhibits a consistent relationship between its time period and orbital radius. According to Kepler's third law, the square of the satellite's orbital period is directly proportional to the cube of its orbital radius. This meanRead more
A satellite orbiting Earth with a specific orbital radius and time period exhibits a consistent relationship between its time period and orbital radius. According to Kepler’s third law, the square of the satellite’s orbital period is directly proportional to the cube of its orbital radius. This means that the ratio of the square of the time period to the cube of the radius remains constant for any satellite orbiting the same central body, such as Earth. This principle reflects the uniformity of gravitational influence and orbital mechanics in determining the motion of satellites around a planet.
When the central gravitational force decreases, it does not produce any torque on the orbiting body because the force acts along the radius vector. As a result, the angular momentum of the body remains conserved. Since angular momentum is directly related to areal velocity (the area swept per unit tRead more
When the central gravitational force decreases, it does not produce any torque on the orbiting body because the force acts along the radius vector. As a result, the angular momentum of the body remains conserved. Since angular momentum is directly related to areal velocity (the area swept per unit time), the areal velocity also remains unchanged. This conservation of angular momentum and areal velocity aligns with Kepler’s second law, which states that a planet sweeps out equal areas in equal time intervals, irrespective of changes in the central force magnitude, as long as no external torque is applied.
A satellite in a circular orbit of radius R has an orbital period of 4 hours. For another satellite orbiting the same planet with a radius of 3R, its orbital period can be determined using Kepler's law of periods. The ratio of the orbital periods is proportional to the ratio of their orbital radii rRead more
A satellite in a circular orbit of radius R has an orbital period of 4 hours. For another satellite orbiting the same planet with a radius of 3R, its orbital period can be determined using Kepler’s law of periods. The ratio of the orbital periods is proportional to the ratio of their orbital radii raised to the power of three-halves. Substituting the values, the period ratio is (3 R/R)³/² = √27. Therefore, the period of the second satellite is √27 x 4, or approximately 4√27 hours.
By the principle of conservation of angular momentum, the orbital period ratio of two orbits is proportional to the ratio of their radii raised to the power of three-halves. For a new orbit with a radius reduced to half of the original, the ratio becomes (R₂/R₁)³/²= 1/2√2. Therefore, the new orbitalRead more
By the principle of conservation of angular momentum, the orbital period ratio of two orbits is proportional to the ratio of their radii raised to the power of three-halves. For a new orbit with a radius reduced to half of the original, the ratio becomes (R₂/R₁)³/²= 1/2√2. Therefore, the new orbital period is 1/2√2 x T₁, where T₁ is the original period. If the initial period is 1 year, the new period becomes \1/2√2 years. This reduction indicates that the duration of the year will decrease significantly.
By conservation of angular momentum,
T₂/T₁ = (R₂/R₁)³/² = ((R₁/2)/R₁)³/² = 1/2√2
T₂ = 1/2√2 T₁ = 1/2√2 x 1 year = 1/2√2 year
Hence the duration of the year will become less.
Satellite is revolving around earth. If its height is increased to four times the height of geostationary satellite, what will become its time period?
According to Kepler's law of periods, the ratio of the orbital periods of two planets is related to the ratio of their semi-major axes. Specifically, the ratio of the periods (T₂/T₁) equals the ratio of their radii raised to the power of three-halves. In this case, if the radius of the second planetRead more
According to Kepler’s law of periods, the ratio of the orbital periods of two planets is related to the ratio of their semi-major axes. Specifically, the ratio of the periods (T₂/T₁) equals the ratio of their radii raised to the power of three-halves. In this case, if the radius of the second planet is four times that of the first, the calculation shows that T₂/T₁ equals eight.
Therefore, if the orbital period of the first planet is 1 day, the orbital period of the second planet would be 8 days, demonstrating the significant impact of radius on orbital time.
According to Kepler’s law of periods,
See lessT₂/T₁ = (r₂/r₁)³/² = (4/1)³/² = 8
T₂ = 8T₁= 8 x 1 day = 8 days
A satellite is orbiting around the earth with orbital radius R and time period T. The quantity which remains constant is
A satellite orbiting Earth with a specific orbital radius and time period exhibits a consistent relationship between its time period and orbital radius. According to Kepler's third law, the square of the satellite's orbital period is directly proportional to the cube of its orbital radius. This meanRead more
A satellite orbiting Earth with a specific orbital radius and time period exhibits a consistent relationship between its time period and orbital radius. According to Kepler’s third law, the square of the satellite’s orbital period is directly proportional to the cube of its orbital radius. This means that the ratio of the square of the time period to the cube of the radius remains constant for any satellite orbiting the same central body, such as Earth. This principle reflects the uniformity of gravitational influence and orbital mechanics in determining the motion of satellites around a planet.
See lessIf gravitational constant is decreasing in time, what will remain unchanged in case of a satellite orbiting around earth?
When the central gravitational force decreases, it does not produce any torque on the orbiting body because the force acts along the radius vector. As a result, the angular momentum of the body remains conserved. Since angular momentum is directly related to areal velocity (the area swept per unit tRead more
When the central gravitational force decreases, it does not produce any torque on the orbiting body because the force acts along the radius vector. As a result, the angular momentum of the body remains conserved. Since angular momentum is directly related to areal velocity (the area swept per unit time), the areal velocity also remains unchanged. This conservation of angular momentum and areal velocity aligns with Kepler’s second law, which states that a planet sweeps out equal areas in equal time intervals, irrespective of changes in the central force magnitude, as long as no external torque is applied.
See lessA satellite in a circular orbit of radius R has a period of 4 h. Another satellite with orbital radius 3 R round the same planet will have a period (in hours)
A satellite in a circular orbit of radius R has an orbital period of 4 hours. For another satellite orbiting the same planet with a radius of 3R, its orbital period can be determined using Kepler's law of periods. The ratio of the orbital periods is proportional to the ratio of their orbital radii rRead more
A satellite in a circular orbit of radius R has an orbital period of 4 hours. For another satellite orbiting the same planet with a radius of 3R, its orbital period can be determined using Kepler’s law of periods. The ratio of the orbital periods is proportional to the ratio of their orbital radii raised to the power of three-halves. Substituting the values, the period ratio is (3 R/R)³/² = √27. Therefore, the period of the second satellite is √27 x 4, or approximately 4√27 hours.
T₂/T₁ = (3 R/R)³/² = √27
See lessT₂ = √27T₁ = 4√27 h.
When the distance between earth and sun is halved, the duration of year will become
By the principle of conservation of angular momentum, the orbital period ratio of two orbits is proportional to the ratio of their radii raised to the power of three-halves. For a new orbit with a radius reduced to half of the original, the ratio becomes (R₂/R₁)³/²= 1/2√2. Therefore, the new orbitalRead more
By the principle of conservation of angular momentum, the orbital period ratio of two orbits is proportional to the ratio of their radii raised to the power of three-halves. For a new orbit with a radius reduced to half of the original, the ratio becomes (R₂/R₁)³/²= 1/2√2. Therefore, the new orbital period is 1/2√2 x T₁, where T₁ is the original period. If the initial period is 1 year, the new period becomes \1/2√2 years. This reduction indicates that the duration of the year will decrease significantly.
By conservation of angular momentum,
See lessT₂/T₁ = (R₂/R₁)³/² = ((R₁/2)/R₁)³/² = 1/2√2
T₂ = 1/2√2 T₁ = 1/2√2 x 1 year = 1/2√2 year
Hence the duration of the year will become less.