Question

(a) Calculate the magnitude of the acceleration due to gravity on the surface of Earth (in m/s2) due to the Moon.
(b) Calculate the magnitude of the acceleration due to gravity at Earth (in m/s2) due to the Sun.
(c) Find the ratio of the Moon's acceleration to the Sun's acceleration.

Answer 1

a) The acceleration due to gravity on the Earth's surface due to the Moon is $3.29\cdot 10^{-5} m/s^2$

b) The acceleration due to gravity on the Earth's surface due to the Sun is $5.88\cdot 10^{-3} m/s^2$

c) The ratio between the Moon's acceleration and the Sun's acceleration is $5.6\cdot 10^{-3}$

Explanation:

a)

The acceleration due to gravity at a certain distance from the centre of a planet is given by

$g=\frac{GM}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

M is the mass of the planet

r is the distance from the centre of the planet

Here we want to calculate the acceleration due to gravity on the surface of Earth due to the Moon. So we have to use:

$M=7.35\cdot 10^{22} kg$, the mass of the Moon

$r=384,400 km + 1737 km = 386,137 km = 3.861 \cdot 10^8 m$ is the distance between the surface of the Earth and the centre of the Moon (we added the radius of the Moon, 1737 km, to the Earth-Moon distance, 384,400 km)

So the acceleration of gravity is

$g_m=\frac{(6.67\cdot 10^{-11})(7.35\cdot 10^{22})}{(3.861\cdot 10^8)^2}=3.29\cdot 10^{-5} m/s^2$

b)

Here instead we want to calculate the acceleration due to gravity on the surface of Earth due to the Sun. So we have to use:

$M=1.99\cdot 10^{30} kg$, the mass of the Sun

$r=149,600,000 km + 695,510 km = 1.503 \cdot 10^8 km=1.503\cdot 10^{11} m$ is the distance between the surface of the Earth and the centre of the Sun (we added the radius of the Sun, 695,510 km, to the Earth-Sun distance, 149,600,000 km)

So the acceleration of gravity is

$g_s=\frac{(6.67\cdot 10^{-11})(1.99\cdot 10^{30})}{(1.503\cdot 10^11)^2}=5.88\cdot 10^{-3} m/s^2$

c)

We have:

Moon's acceleration:

$g_m = 3.29\cdot 10^{-5} m/s^2$

Sun's acceleration:

$g_s=5.88\cdot 10^{-3} m/s^2$

So the ratio between the Moon's acceleration to the Sun's acceleration is:

$\frac{g_m}{g_s}=\frac{3.29\cdot 10^{-5}}{5.88\cdot 10^{-3}}=5.6\cdot 10^{-3}$

Learn more about gravity here:

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Answer 2

Answer:

$g_{moon}=3.43*10^{-5}m/s^{2}$.

$g_{sun}=5.93*10^{-3}m/s^{2}$.

$\frac{g_{moon}}{g_{sun}}=5.78*10^{-3}$.

Explanation:

For this problem, we are going to use Newton's law of universal gravitation, which states that two bodies attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The law is defined as:

$F_{g}=\frac{GMm}{r^{2}}$,

where M and m are the masses of the objects, G is gravitational constant and r the distances between the centers of mass of the bodies.

To find the acceleration due to gravitational of a massive body we have

$F_{g}=\frac{GMm}{r^{2}}$,

and from Newton's Second Law we also know that

$F_g=ma$ so,

$ma=\frac{GMm}{r^{2}}$,

$a=\frac{GM}{r^{2}}$ (we are going to call this equation eq1),

here $a$ is the acceleration due to a massive body (the gravity due to that body!).

A)

For computing the gravity due to the moon we need to remember that

• moon's mass is $M_{moon}=7.35*10^{22}kg$ and,

• moon's distance from earth's surface is $r=378029 km$.

Now we can easily compute moon's gravity on earth surface using eq1:

$g_{moon}=\frac{GM}{r^{2}}$,

$g_{moon}=\frac{(6.67*10^{-11})(7.35*10^{22})}{(378029*10^{3})^{2}}$,

$g_{moon}=3.43*10^{-5}m/s^{2}$.

B)

We are going to proceed the same as before,

• sun's mass is $M_{sun}=1.99^{30}kg$ and,
• sun's distance from earth's surface is $r=149593629 km$.

Computing sun's gravity on earth surface using eq1:

$g_{sun}=\frac{GM}{r^{2}}$,

$g_{sun}=\frac{(6.67*10^{-11})(1.99*10^{30})}{(149593629*10^{3})^{2}}$,

$g_{sun}=5.93*10^{-3}m/s^{2}$.

C)

The ratio is simply

$\frac{g_{moon}}{g_{sun}}=\frac{3.43*10^{-5}}{5.93*10^{-3}}$

$\frac{g_{moon}}{g_{sun}}=5.78*10^{-3}$,

which means that acceleration due to the sun is much more greater!