Suppose one night the radius of the earth doubled but its mass stayed the same. What would be an approximate new value for the free-fall acceleration at the surface of the earth?

Answer: To find the new value for the free-fall acceleration at the surface of the Earth when the radius doubles but the mass stays the same, we can use the formula for gravitational acceleration:

g = G * (M / r^2)

where:

- g is the free-fall acceleration,

- G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2),

- M is the mass of the Earth, and

- r is the radius of the Earth.

If the radius of the Earth doubles, we can denote the original radius as r₀ and the new radius as 2r₀.

Since the mass of the Earth stays the same, we can denote it as M.

Using these values, we can calculate the new free-fall acceleration:

g' = G * (M / (2r₀)^2)

Simplifying:

g' = G * (M / 4r₀^2)

Dividing both the numerator and denominator by 4:

g' = (G / 4) * (M / r₀^2)

As we can see, the new free-fall acceleration (g') is (G / 4) times the original free-fall acceleration (g).

Therefore, if the radius of the Earth doubles but its mass stays the same, the approximate new value for the free-fall acceleration at the surface of the Earth would be one-fourth (1/4) of the original value.

Explanation: you barely gave me information

msm555 msm555 · Answer 2 · 2024-01-06T08:16:54+01:00

Answer:

[tex]\sf A. \, 2.5 \, \textsf{m/s}^2[/tex].

Explanation:

The formula gives the gravitational acceleration at the surface of the Earth:

[tex]\sf \boxed{\boxed{g = \dfrac{GM}{R^2}}} [/tex]

where:

[tex]\sf G [/tex] is the gravitational constant ([tex]\sf 6.67430 \times 10^{-11} \, \textsf{m}^3 \, \textsf{kg}^{-1} \, \textsf{s}^{-2}[/tex]),
[tex]\sf M [/tex] is the mass of the Earth ([tex]\sf 5.972 \times 10^{24} \, \textsf{kg}[/tex]),
[tex]\sf R [/tex] is the radius of the Earth.

If the radius ([tex]\sf R [/tex]) of the Earth doubled, the new radius ([tex]\sf R' [/tex]) would be [tex]\sf 2R[/tex].

Substituting this into the formula, we get:

[tex]\sf g' = \dfrac{GM}{(2R)^2} [/tex]

Simplifying this expression:

[tex]\sf g' = \dfrac{GM}{4R^2} [/tex]

Now, we can express [tex]\sf g' [/tex] in terms of the original gravitational acceleration [tex]\sf g [/tex]:

[tex]\sf g' = \dfrac{1}{4} g [/tex]

So, if the radius of the Earth doubled but its mass stayed the same, the new free-fall acceleration ([tex]\sf g' [/tex]) at the surface of the Earth would be approximately one-fourth ([tex]\sf \dfrac{1}{4} [/tex]) of the original gravitational acceleration ([tex]\sf g [/tex]).

If we take the approximate value of [tex]\sf g [/tex] as [tex]\sf 9.8 \, \textsf{m/s}^2[/tex], then the new value for [tex]\sf g' [/tex] would be:

[tex]\sf g' \approx \dfrac{1}{4} \times 9.8 \, \textsf{m/s}^2\\\\ = 2.45 \, \textsf{m/s}^2\\\\ 2.5 \, \textsf{m/s}^2 \\\\ \textsf{(in nearest whole number)} [/tex]

So, the new approximate value for the free-fall acceleration at the surface of the Earth would be [tex]\sf A. \, 2.5 \, \textsf{m/s}^2[/tex].