Question:
We can use two mathematical models to calculate the work done when a 300 kg satellite is moved from Earth’s surface to an altitude of 200 km. Model A uses the mathematical equation, W = mg∆h, whereas Model B uses the mathematical equation, ΔE = −GMm/(R + h) – (–GMm/R).
(a) State the assumptions made on Earth’s gravitational field in models X and Y.
(b) Explain why models X and Y produce results that are approximately the same.
We can use two mathematical models to calculate the work done when a 300 kg satellite is moved from Earth’s surface to an altitude of 200 km. Model A uses the mathematical equation, W = mg∆h, whereas Model B uses the mathematical equation, ΔE = −GMm/(R + h) – (–GMm/R).
(a) State the assumptions made on Earth’s gravitational field in models X and Y.
(b) Explain why models X and Y produce results that are approximately the same.
Marking Guidelines
Question 26 (a) Criteria
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Marks
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• Identifies correct assumptions.
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2
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• Identifies an assumption with correct information.
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1
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Possible answer:
Assumption for Model X: Earth’s gravitational field is uniform.
Assumption for Model X: Earth’s gravitational field is uniform.
Assumption for Model Y: Earth’s gravitational field varies with altitude.
Question 26 (b) Criteria
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Mark
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• Correct reason.
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1
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Possible answer: Variations in Earth’s gravitational field from the surface to an altitude of 200 km are sufficiently small.
(Source: https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/guides/2015-hsc-mg-physics.pdf)
Comments:
In part (a), possible answer includes “the Earth’s gravitational field is uniform.” Alternatively, students might answer that the gravitational field is approximately constant on the Earth’s surface and up to an altitude of 200 km (as specified in the question). That is, we may assume the gravitational field strength to be constant (about 9.8 N/kg) if the satellite is not going beyond a certain height. Therefore, the work done in lifting the satellite can be mathematically expressed by the equation, W = mg∆h, because the gravitational force (mg) is assumed to be constant from Earth’s surface to the altitude (∆h) of 200 km. However, this assumption does not always hold for a low Earth orbit satellite which may have an altitude between 160 km and 2,000 km.
On the other hand, the sample answer states that “Model Y assumes the gravitational field changes with altitude.” To be precise, students could answer that the gravitational field follows the inverse square law (g = GM/r2). However, some students might include additional assumptions such as the density of the Earth is constant, the Earth is not rotating, and the Earth is spherical. Strictly speaking, the gravitational field is not even constant or the same at different locations on the Earth’s surface.
In part (b), the sample answer states that “variations in gravitational attraction from the Earth’s surface to an altitude of 200 km are sufficiently small.” However, 200 km is still quite a long distance. Thus, we should explain that the altitude of 200 km is negligible if it is compared to the radius of the Earth which is about 6,370 km. We can clarify this fact mathematically as shown below:
Change in gravitational potential energy, ΔE = −GMm/(R + h) – (–GMm/R)
= (−GMm/R) (1 + h/R)–1 – (–GMm/R) ≈ (−GMm/R) (1 – h/R – 1)
= (GMmh/R2) = mgh
Note: By using (1 + x)–1 = 1 – x for small values of x.
The approximation is appropriate because the altitude of 200 km is relatively short as compared to the radius of the Earth (i.e. h/R << 1). However, we can also explain that the error in work done increases as the altitude increases. The error is due to the neglected terms such as (−GMm/R) (h/R)2.
Note: (1 + x)–1 = 1 – x + x2 – x3 + …
Feynman’s insights or goofs?:
Interestingly, Feynman explains that “[i]f we have a gravitational field that is uniform, if we are not going to heights comparable with the radius of the earth, then the force is a constant vertical force and the work done is simply the force times the vertical distance (Feynman et al., 1963, section 14-3 Conservative forces).” That is, the work done lifting an object is equal to the gravitational force times the vertical distance of the object raised. However, this simplification is possible only if the object is not moving to heights that are comparable to the radius of the earth. In other words, if the height increases, the error in calculating the work done increases.
More importantly, in Feynman’s words, “[t]he gravitational field of the earth is not precisely uniform, so a freely falling ball has a slightly different acceleration at different places — the direction changes and the magnitude changes (Feynman et al., 1964, section 42-5 Gravity and the principle of equivalence).” Simply put, the gravitational field of the Earth is not strictly constant and it can vary depending on the location of the Moon and the Sun. Essentially, the gravitational field of the Moon can affect the shape of the Earth and this further change the gravitational field of the Earth.
In part (a), possible answer includes “the Earth’s gravitational field is uniform.” Alternatively, students might answer that the gravitational field is approximately constant on the Earth’s surface and up to an altitude of 200 km (as specified in the question). That is, we may assume the gravitational field strength to be constant (about 9.8 N/kg) if the satellite is not going beyond a certain height. Therefore, the work done in lifting the satellite can be mathematically expressed by the equation, W = mg∆h, because the gravitational force (mg) is assumed to be constant from Earth’s surface to the altitude (∆h) of 200 km. However, this assumption does not always hold for a low Earth orbit satellite which may have an altitude between 160 km and 2,000 km.
On the other hand, the sample answer states that “Model Y assumes the gravitational field changes with altitude.” To be precise, students could answer that the gravitational field follows the inverse square law (g = GM/r2). However, some students might include additional assumptions such as the density of the Earth is constant, the Earth is not rotating, and the Earth is spherical. Strictly speaking, the gravitational field is not even constant or the same at different locations on the Earth’s surface.
In part (b), the sample answer states that “variations in gravitational attraction from the Earth’s surface to an altitude of 200 km are sufficiently small.” However, 200 km is still quite a long distance. Thus, we should explain that the altitude of 200 km is negligible if it is compared to the radius of the Earth which is about 6,370 km. We can clarify this fact mathematically as shown below:
Change in gravitational potential energy, ΔE = −GMm/(R + h) – (–GMm/R)
= (−GMm/R) (1 + h/R)–1 – (–GMm/R) ≈ (−GMm/R) (1 – h/R – 1)
= (GMmh/R2) = mgh
Note: By using (1 + x)–1 = 1 – x for small values of x.
The approximation is appropriate because the altitude of 200 km is relatively short as compared to the radius of the Earth (i.e. h/R << 1). However, we can also explain that the error in work done increases as the altitude increases. The error is due to the neglected terms such as (−GMm/R) (h/R)2.
Note: (1 + x)–1 = 1 – x + x2 – x3 + …
Feynman’s insights or goofs?:
Interestingly, Feynman explains that “[i]f we have a gravitational field that is uniform, if we are not going to heights comparable with the radius of the earth, then the force is a constant vertical force and the work done is simply the force times the vertical distance (Feynman et al., 1963, section 14-3 Conservative forces).” That is, the work done lifting an object is equal to the gravitational force times the vertical distance of the object raised. However, this simplification is possible only if the object is not moving to heights that are comparable to the radius of the earth. In other words, if the height increases, the error in calculating the work done increases.
More importantly, in Feynman’s words, “[t]he gravitational field of the earth is not precisely uniform, so a freely falling ball has a slightly different acceleration at different places — the direction changes and the magnitude changes (Feynman et al., 1964, section 42-5 Gravity and the principle of equivalence).” Simply put, the gravitational field of the Earth is not strictly constant and it can vary depending on the location of the Moon and the Sun. Essentially, the gravitational field of the Moon can affect the shape of the Earth and this further change the gravitational field of the Earth.
References:
1. Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.
2. Feynman, R. P., Leighton, R. B., & Sands, M. L. (1964). The Feynman Lectures on Physics, Vol II: Reading, MA: Addison-Wesley.
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