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.

Question

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

Answer 1

To find the acceleration due to gravity on the surface of a shrunken Earth, we can use the formula for gravitational acceleration:

𝑔 = 𝐺 × (𝑀/𝑟^2)

where 𝑔 is the acceleration due to gravity, 𝐺 is the gravitational constant, 𝑀 is the mass of Earth, and 𝑟 is the radius of Earth.

Given that the mass of Earth and its shape remain the same, we can simply plug in the new value of the radius into the formula to find the new acceleration due to gravity.

Let's calculate:

New radius (𝑟𝑛𝑒𝑤) = 0.21 × original radius
𝑟𝑛𝑒𝑤 = 0.21 × 6378.1 km (original radius of Earth)

Now we can substitute the values into the formula:

𝑔𝑛𝑒𝑤 = 𝐺 × (𝑀/𝑟𝑛𝑒𝑤^2)

We know that 𝐺 = 9.80 m/s^2 (gravitational constant) and 𝑀 is the mass of Earth, which remains the same. So we'll just focus on the new radius:

𝑟𝑛𝑒𝑤 = 0.21 × 6378.1 km = 1341.9821 km

Converting the radius into meters:

𝑟𝑛𝑒𝑤 = 1341.9821 km × 1000 m/km = 1341982.1 m

Substituting the values into the formula:

𝑔𝑛𝑒𝑤 = 9.80 × (𝑀/𝑟𝑛𝑒𝑤^2) = 9.80 × (5.972 × 10^24 kg) / (1341982.1 m)^2

Calculating the value:

𝑔𝑛𝑒𝑤 ≈ 2.97 m/s^2

Therefore, 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, the acceleration due to gravity on its surface will be approximately 2.97 m/s^2.