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A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius (1.01)R. The period of the second satellite is larger than that of the first one by approximately [IIT 1995]
- a)1.5%
- b)1.0%
- c)0.5%
- d)3.0%
Correct answer is option 'A'. Can you explain this answer?
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A satellite is launched into a circular orbit of radius R around the e...
To understand why the period of the second satellite is larger than that of the first one, we need to consider the relationship between the period and the radius of the orbit.
1. Understanding Orbital Period:
The period of an object in a circular orbit is the time it takes for the object to complete one full revolution around the central body. In the case of satellites orbiting the Earth, the period is the time it takes for the satellite to complete one orbit around the Earth.
2. Kepler's Third Law:
Kepler's Third Law states that the square of the orbital period of a planet (or satellite) is directly proportional to the cube of its average distance from the central body. Mathematically, it can be expressed as:
T^2 ∝ R^3
where T is the period and R is the radius of the orbit.
3. Comparing the Two Satellites:
Let's consider the first satellite with a radius of R and a period of T. According to Kepler's Third Law:
T^2 ∝ R^3
Now, let's consider the second satellite with a radius of (1.01)R. We want to find the period of this satellite, which we'll call T'.
T'^2 ∝ (1.01R)^3
T'^2 ∝ 1.01^3 * R^3
T'^2 ∝ 1.030301 * R^3
Comparing the two equations, we can see that the period of the second satellite is larger than that of the first one by a factor of approximately 1.030301. To find the percentage increase, we can subtract 1 from this factor and multiply by 100:
Percentage Increase = (1.030301 - 1) * 100
Percentage Increase ≈ 3.03%
However, the options provided in the question are given in increments of 0.5%. To find the closest option, we can round the percentage increase to the nearest 0.5%:
Rounded Percentage Increase ≈ 3.0%
Therefore, the correct answer is option (d) 3.0%.
1. Understanding Orbital Period:
The period of an object in a circular orbit is the time it takes for the object to complete one full revolution around the central body. In the case of satellites orbiting the Earth, the period is the time it takes for the satellite to complete one orbit around the Earth.
2. Kepler's Third Law:
Kepler's Third Law states that the square of the orbital period of a planet (or satellite) is directly proportional to the cube of its average distance from the central body. Mathematically, it can be expressed as:
T^2 ∝ R^3
where T is the period and R is the radius of the orbit.
3. Comparing the Two Satellites:
Let's consider the first satellite with a radius of R and a period of T. According to Kepler's Third Law:
T^2 ∝ R^3
Now, let's consider the second satellite with a radius of (1.01)R. We want to find the period of this satellite, which we'll call T'.
T'^2 ∝ (1.01R)^3
T'^2 ∝ 1.01^3 * R^3
T'^2 ∝ 1.030301 * R^3
Comparing the two equations, we can see that the period of the second satellite is larger than that of the first one by a factor of approximately 1.030301. To find the percentage increase, we can subtract 1 from this factor and multiply by 100:
Percentage Increase = (1.030301 - 1) * 100
Percentage Increase ≈ 3.03%
However, the options provided in the question are given in increments of 0.5%. To find the closest option, we can round the percentage increase to the nearest 0.5%:
Rounded Percentage Increase ≈ 3.0%
Therefore, the correct answer is option (d) 3.0%.
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A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius (1.01)R. The period of the second satellite is larger than that of the first one by approximately [IIT 1995]a)1.5%b)1.0%c)0.5%d)3.0%Correct answer is option 'A'. Can you explain this answer?
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