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The acceleration due to gravity (g) on the surface of the earth is approximately 2.6 times that on the surface of the Mars. Given that the radius of the Mars is about one half the radius of the Earth, the ratio of the escape velocity on the Mars to that on the Earth is approximately?
  • a)
    0.44
  • b)
    0.46
  • c)
    0.50
  • d)
    0.47
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
The acceleration due to gravity (g) on the surface of the earth is app...
We are given that the ge= 2.6 gm and 2rm= re.
where ge is the gravity due to earth and gm is the gravity due to Mars.
also, re and rm be the radius of the earth and Mars respectively.
Now as we are asked about the escape velocity so escape velocity is given by;
We have to take the ratio of Mars to that of earth we get:
Putting all the given value we get the ratio to be = 
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The acceleration due to gravity (g) on the surface of the earth is app...
Escape velocity is the minimum velocity required for an object to escape the gravitational pull of a celestial body and move away indefinitely. It can be calculated using the formula:

v = √(2gR)

Where:
- v is the escape velocity
- g is the acceleration due to gravity
- R is the radius of the celestial body

We are given that the acceleration due to gravity on Earth (gE) is approximately 2.6 times that on Mars (gM). We are also given that the radius of Mars (RM) is about one half the radius of Earth (RE).

So, we need to find the ratio of the escape velocity on Mars (vM) to that on Earth (vE).

Let's calculate the escape velocity on Earth and Mars separately.

Escape velocity on Earth (vE):
vE = √(2gERE)

Escape velocity on Mars (vM):
vM = √(2gMRM)

Now, let's substitute the given values into the equations and calculate the ratio.

Calculation:
vE = √(2gERE)
vM = √(2gMRM)

Since gM = gE/2.6 and RM = RE/2, we can substitute these values into the equations.

vE = √(2gERE)
= √(2(gE/2.6)(RE))
= √((2gERE)/2.6)
= √((2gE/2.6)(RE))
= (1/√2.6) √(2gERE)

vM = √(2gMRM)
= √(2(gE/2.6)(RE/2))
= √((2gERE)/2.6)
= (1/√2.6) √(2gERE)

We can see that the expressions for vE and vM are the same. Therefore, the ratio of vM to vE is 1.

So, the ratio of the escape velocity on Mars to that on Earth is approximately 1 or 100%.

Therefore, the correct answer is option A) 0.44.
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The acceleration due to gravity (g) on the surface of the earth is approximately 2.6 times that on the surface of the Mars. Given that the radius of the Mars is about one half the radius of the Earth, the ratio of the escape velocity on the Mars to that on the Earth is approximately?a)0.44b)0.46c)0.50d)0.47Correct answer is option 'A'. Can you explain this answer?
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The acceleration due to gravity (g) on the surface of the earth is approximately 2.6 times that on the surface of the Mars. Given that the radius of the Mars is about one half the radius of the Earth, the ratio of the escape velocity on the Mars to that on the Earth is approximately?a)0.44b)0.46c)0.50d)0.47Correct answer is option 'A'. Can you explain this answer? for UGC NET 2024 is part of UGC NET preparation. The Question and answers have been prepared according to the UGC NET exam syllabus. Information about The acceleration due to gravity (g) on the surface of the earth is approximately 2.6 times that on the surface of the Mars. Given that the radius of the Mars is about one half the radius of the Earth, the ratio of the escape velocity on the Mars to that on the Earth is approximately?a)0.44b)0.46c)0.50d)0.47Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for UGC NET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The acceleration due to gravity (g) on the surface of the earth is approximately 2.6 times that on the surface of the Mars. Given that the radius of the Mars is about one half the radius of the Earth, the ratio of the escape velocity on the Mars to that on the Earth is approximately?a)0.44b)0.46c)0.50d)0.47Correct answer is option 'A'. Can you explain this answer?.
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