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Since the answer isn't obvious to you, it's probably best that we take what may seem like a very long way to find the answer. Let's start by looking at two equations: 1) for the force of gravity; and, 2) one that relates force to acceleration.
1) F = (GMm)/r^2, where
G = Universal gravitational constant, 6.67X10^-11 Nt-m^2/s^2
M = Mass of one body (the planet in this case)
m = Mass of the other body (any object at the surface of that planet)
r = Radius of the planet
2) F = ma, where in this case
F = Force of attraction between the planet and some object
m = Mass of that object
a = Acceleration of that object
Since they want the acceleration due to gravity, let's look at the general case. We can find the force of gravity from equation 1, and we can relate that force to an acceleration by equation 2. The F is the same in both equations, so we can set them equal to each other and get
ma = (GMm)/r^2, and the little m's disappear, leaving
a = (GM)/r^2, which is 1g
That gives us the acceleration, which is independent of the mass of the small object. In the case of the Earth, you could substitute in the values for G, M and r for Earth and solve for 'a', and you would find that it works out to be 9.81m/s^2. That is 1g, the acceleration that we experience every day. So we have that
1g = (GM)/r^2 , when
M = mass of the Earth
r = radius of the Earth
Now, let's put in the values for Planet X and see how they compare to the Earth. The mass is 5.34 times that of the Earth (5.34M), and the radius is 4.35 times the radius of Earth (4.35r). The equation then becomes that
a = (G X 5.34M)/(4.35r)^2
a = (5.34)(GM)/(18.92)(r^2)
a = (5.34/18.92) X (GM)/r^2, numbers are rounded a bit
a = 0.282 X (GM)/r^2
Since (GM)/r^2 = 1g, we can make that substitution and get
a = 0.282g
I hope that didn't make it even MORE confusing.
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