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Fish181
Joined: 13 Dec 2023
Last visit: 22 Jan 2025
Posts: 133
Given Kudos: 53
Status:Applying in R1 of 2024 to t15
Affiliations: University of Tennessee
Location: United States (CO)
Concentration: Strategy, Finance
GMAT Focus 1: 605 Q76 V84 DI80 
GMAT Focus 2: 615 Q78 V86 DI78
GPA: 3.62
WE:Analyst (Consumer Packaged Goods)
GMAT Focus 2: 615 Q78 V86 DI78
Posts: 133
Updated on: 18 Mar 2024, 12:49
Originally posted by Fish181 on 18 Mar 2024, 09:18.
Last edited by gmatophobia on 18 Mar 2024, 12:49, edited 1 time in total.
Last edited by gmatophobia on 18 Mar 2024, 12:49, edited 1 time in total.
formatted the question and corrected the tags
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0.5 km: 3km
1.5 km: 9km
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
78% (02:29) correct
22%
(02:22)
wrong
based on 597
sessions
History
Date
Time
Result
Not Attempted Yet
A satellite is currently in a circular orbit with radius 32,714 km about the center of the earth.
Select for 0.5 km the increase, to the nearest whole kilometer, in the distance the satellite travels about the center of the earth during each revolution if the orbital radius of the satellite is increased by 0.5 km. And select for 1.5 km the increase, to the nearest whole kilometer, in the distance the satellite travels about the center of the earth during each revolution if the orbital radius of the satellite is increased by 1.5 km. Make only two selections, one in each column.
Select for 0.5 km the increase, to the nearest whole kilometer, in the distance the satellite travels about the center of the earth during each revolution if the orbital radius of the satellite is increased by 0.5 km. And select for 1.5 km the increase, to the nearest whole kilometer, in the distance the satellite travels about the center of the earth during each revolution if the orbital radius of the satellite is increased by 1.5 km. Make only two selections, one in each column.
| 0.5 km | 1.5 km | |
| 3km | ||
| 9km | ||
| 27km | ||
| 81km | ||
| 150km | ||
| 450km |
ShowHide Answer
Official Answer
0.5 km: 3km
1.5 km: 9km
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29 Mar 2024, 01:01
Kudos
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A satellite is currently in a circular orbit with radius 32,714 km about the center of the earth.
Select for 0.5 km the increase, to the nearest whole kilometer, in the distance the satellite travels about the center of the earth during each revolution if the orbital radius of the satellite is increased by 0.5 km. And select for 1.5 km the increase, to the nearest whole kilometer, in the distance the satellite travels about the center of the earth during each revolution if the orbital radius of the satellite is increased by 1.5 km. Make only two selections, one in each column.
According to GMAC, Geometry does not appear on the Focus Edition, but questions, such as this one, on the official practice tests indicate that knowing some basic Geometry concepts that can be considered common knowledge is necessary for getting some Quant questions correct.
Among those concepts are the Pythagorean theorem and formulas for calculating the areas of basic shapes including rectangles, triangles, and circles.
Now, this question requires use of the formula for the circumference of a circle, which is \(C = 2πr = πd\). where \(C\) is the circumference, \(r\) is the radius, \(d\) is the diameter of the circle, and the value of \(π\) is approximately \(3.14\).
Given that formula, we can see that, any time we increase the radius of a circle, we increase the circumference of the circle by \(2π\) times the increase in the radius, or by about \(6.28\) times the increase in the radius.
So, to calculate the increase in the distance the satellite travels during each revolution, we can ignore the figure 32,714 and just multiply each increase in the radius by 6.28.
If the radius is increased by 0.5 km, the distance is increased by approximately 0.5 × 6.28 = 3.14, which rounded to the nearest whole kilometer is 3.
If the radius is increased by 1.5 km, the distance is increased by approximately 1.5 × 6.28 = 9.42, which rounded to the nearest whole kilometer is 9.
3km
9km
27km
81km
150km
450km
Correct answer: 3km, 9km
Select for 0.5 km the increase, to the nearest whole kilometer, in the distance the satellite travels about the center of the earth during each revolution if the orbital radius of the satellite is increased by 0.5 km. And select for 1.5 km the increase, to the nearest whole kilometer, in the distance the satellite travels about the center of the earth during each revolution if the orbital radius of the satellite is increased by 1.5 km. Make only two selections, one in each column.
According to GMAC, Geometry does not appear on the Focus Edition, but questions, such as this one, on the official practice tests indicate that knowing some basic Geometry concepts that can be considered common knowledge is necessary for getting some Quant questions correct.
Among those concepts are the Pythagorean theorem and formulas for calculating the areas of basic shapes including rectangles, triangles, and circles.
Now, this question requires use of the formula for the circumference of a circle, which is \(C = 2πr = πd\). where \(C\) is the circumference, \(r\) is the radius, \(d\) is the diameter of the circle, and the value of \(π\) is approximately \(3.14\).
Given that formula, we can see that, any time we increase the radius of a circle, we increase the circumference of the circle by \(2π\) times the increase in the radius, or by about \(6.28\) times the increase in the radius.
So, to calculate the increase in the distance the satellite travels during each revolution, we can ignore the figure 32,714 and just multiply each increase in the radius by 6.28.
If the radius is increased by 0.5 km, the distance is increased by approximately 0.5 × 6.28 = 3.14, which rounded to the nearest whole kilometer is 3.
If the radius is increased by 1.5 km, the distance is increased by approximately 1.5 × 6.28 = 9.42, which rounded to the nearest whole kilometer is 9.
3km
9km
27km
81km
150km
450km
Correct answer: 3km, 9km
General Discussion
18 Mar 2024, 11:55
Kudos
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The path distance can be represented by:
2pi(R+r)
where R is radius of Earth and r height of satellite above Earth.
R and pi don't change, so the the change in the length of its orbit is solely due to the change in r.
Therefore, with a change in r of 0.5 and 1.5, the path changes by:
2pi*.5 and 2pi*1.5, or:
~3 and ~9, respectively.
Posted from my mobile device
2pi(R+r)
where R is radius of Earth and r height of satellite above Earth.
R and pi don't change, so the the change in the length of its orbit is solely due to the change in r.
Therefore, with a change in r of 0.5 and 1.5, the path changes by:
2pi*.5 and 2pi*1.5, or:
~3 and ~9, respectively.
Posted from my mobile device
