If Earth shrinks in size such that its shape and mass remain the same, but the radius decreases to 0.21 times its original value, find the acceleration due to gravity on its surface.
(6378.1/0.21)(9.80)
= 2.97 m/s^2
No. The inverse law applies. If the distance shrinks, the force increases.
g= 9.8/(1/.21)^2
not in the answer choices...
a-2.97 m/s^2
b-2.97 x 10^8 m/s^2
c- 220 m/s^2
d- 22x10^8 m/s^2
To find the acceleration due to gravity on the surface of Earth after it shrinks, we can use the formula for gravitational acceleration:
g' = (GM) / (r')^2
Where:
- g' is the acceleration due to gravity on the new Earth surface
- G is the gravitational constant (approximated as 6.67430 × 10^-11 m^3⋅kg^-1⋅s^-2)
- M is the mass of Earth (which remains the same)
- r' is the new radius of Earth
According to the information given, the new radius of Earth (r') is 0.21 times the original radius. Let's assume the original radius is denoted by r.
So, r' = 0.21 * r
Now, we can substitute these values into the formula:
g' = (GM) / (r')^2
= (GM) / ((0.21 * r)^2)
= (GM) / (0.0441 * r^2)
= (1/0.0441) * (GM) / r^2
= (GM) / (0.0441 * r^2)
We know that G is a constant and the mass of Earth (M) remains the same. Therefore, we can simplify the equation:
g' = (GM) / (0.0441 * r^2)
= g / 0.0441
Where g is the acceleration due to gravity on the original Earth surface.
Now, we need to find the value of g. The standard value of g on Earth is approximately 9.80 m/s^2. Substituting this value into the equation:
g' = (9.80) / 0.0441
= 222.168 m/s^2
Therefore, the acceleration due to gravity on the surface of the shrunken Earth is approximately 222.168 m/s^2.
It seems there was an error in your calculation. Please recheck your calculations to find the correct result.