Question

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?
A. 2.5m/s22.5m/s 2
B. 5.0m/s25.0m/s 2
C. 10m/s210m/s 2
D. 20m/s220m/s 2

Answer 1

Answer:

The free-fall acceleration at the surface of the new earth is $a = 2.5 m/s^2$, letter A.

Explanation:

Applying the equivalence principle we have:

Gravitational Force:

$F=G\frac{mM}{R^{2}}$

Newton's second law

$F=ma$

Matching both equations, we have:

$ma=G\frac{mM}{R^{2}}$

$a=G\frac{M}{R^{2}}$

Where:

M is the mass of the earth $5.972*10^{24} kg$

R is the radius of this new earth. In this case double of the earth $(2*6371 km = 12742 km = 1.274*10^{7} m)$

G is the gravitational constant $6.674*10^{-11}Nm^{2}kg^{-2}$

$a=6.674*10^{-11}\frac{5.972*10^{24}}{(1.274*10^{7})^{2}}$

$a=2.45 ms^{-2}$

Therefore, the free-fall acceleration at the surface of the new earth is $a = 2.5 m/s^2$, letter A.

I hope it helps you!

Answer 2

The free-fall acceleration at the surface of the new earth is 2.5m/s².

What is Acceleration?

This is defined as the rate of change of the velocity of an object with respect to time.

Gravitational Force = F = GMm/r²

F = ma

Combine both equations

ma = GMm/r²

a = GM/r²

a = 6.674 ˣ 10⁻¹¹  ˣ (5.972 ˣ 10²⁴) / (1.274ˣ 10⁷)²

a= 2.45m/s²

Therefore the most appropriate choice is option A.

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