proportionality question

[AmlanMishra]

The gravitational pull on an object is directly proportional to its mass, m, and inversely proportional to the square of its distance, d, from the centre of mass of the Earth. If an object is moved to a distance twice as far from the centre of mass of the Earth, what should the new mass of the object be for the gravitational pull on it to remain unchanged, in terms of m?

A- m / 4
B- m /2
C - 2m
D- 4m
E - root8 m

[GMATGuruNY]

The gravitational pull on an object is directly proportional to its mass, m, and inversely proportional to the square of its distance, d, from the centre of mass of the Earth. If an object is moved to a distance twice as far from the centre of mass of the Earth, what should the new mass of the object be for the gravitational pull on it to remain unchanged, in terms of m?

A- m / 4
B- m /2
C - 2m
D- 4m
E - root8 m

The gravitational pull on an object is directly proportional to its mass, m, and inversely proportional to the square of its distance, d, from the centre of mass of the Earth.
Let p = pull, m = mass, d = distance.
The statement above, translated into math:
p = k(m/d²), where k is a constant.
In the blue equation:
If m doubles, then p doubles, with the result that m and p are DIRECTLY PROPORTIONAL.
If d² doubles, then p is cut in half, with the result that d² and p are INVERSELY PROPORTIONAL.

Let the original values be as follows:
m=1, d=1, k=1.
Plugging these values into the blue equation, we get:
p = 1(1/1²) = 1.

An object is moved to a distance twice as far from the centre of mass of the Earth.
The gravitational pull on it to remain unchanged.
New values:
Since p is unchanged, p=1.
Since d doubles, d=2.
Since k is a constant, k=1.
Let n = the new mass.
Plugging these values into the blue equation, we get:
1 = 1(n/2²)
n = 4.

What should the new mass of the object be?
Since n=4, this is our target.
Now plug m=1 into the answer choices to see which yields our target of 4.
Only D works:
4m = 4*1 = 4.

The correct answer is D.

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