Question

Assume that the acceleration due to gravity on the surface of the moon is 0.2 times the acceleration due to gravity on the surface of earth. If ${R_e}$ is the maximum range of a projectile on the earth’s surface. What is the maximum range on the surface of the moon for the same velocity of projection?
A) $0.2{R_e}$
B) $2{R_e}$
C) $0.5{R_e}$
D) $5{R_e}$

Answer 1

Projectile motion is a type of motion in which the object’s body is thrown into the air and is subjected to the acceleration due to gravity only. Here, imagine you throwing a ball upward in the air on Earth and with the same velocity throwing the ball upward on the moon. Equate both of the projectiles' equations together and solve.

Complete step by step solution:
Find the maximum range on the surface of the moon for the same velocity.
Here, the formula for projectile motion:
$R = \dfrac{{v_o^2\sin 2{\theta _o}}}{g}$;
Where:
R = Range of the projectile.
${v_o}$= Velocity of the projectile.
g = Acceleration due to gravity.

Now, write the formula for projectile motion for Earth.
${R_e} = \dfrac{{v_o^2\sin 2{\theta _o}}}{{{g_e}}}$; …(1)
Now, Similarly for the Moon,
${R_m} = \dfrac{{v_o^2\sin 2{\theta _o}}}{{{g_m}}}$; …(2)
Now, we have been given that acceleration due to gravity on surface of the moon is 0.2 times the acceleration due to gravity on the surface of earth:
${g_m} = 0.2{g_e}$; …(3)
Now,
${R_e} = \dfrac{{v_o^2\sin 2{\theta _o}}}{{{g_e}}}$;
$\Rightarrow {R_e} \times {g_e} = v_o^2\sin 2{\theta _o}$; …(5)
Similarly,
${R_m} = \dfrac{{v_o^2\sin 2{\theta _o}}}{{{g_m}}}$;
$\Rightarrow {R_m} \times {g_m} = v_o^2\sin 2{\theta _o}$; …(6)
Equate equations 5 and 6 together:
${R_m} \times {g_m} = {R_e} \times {g_e}$ ….(7)
Now we know that ${g_m} = 0.2{g_e}$. Put it in the above equation:
${R_m} \times 0.2{g_e} = {R_e} \times {g_e}$; …(8)
Cancel out the common factor:
$\Rightarrow {R_m} \times 0.2 = {R_e}$; …(9)
Write the above equation in terms of ${R_m}$:
$\Rightarrow {R_m} = \dfrac{{{R_e}}}{{0.2}}$;
Do, the necessary calculation:
$\Rightarrow {R_m} = 5{R_e}$;

Final Answer: Option “D” is correct. The maximum range on the surface of the moon for the same velocity of projection is ${R_m} = 5{R_e}$.

Note: Write the equation for the projectile motion for Earth and the moon. We have been given the same velocity of projection, equate the equation for projectile motion for both Earth and Moon together and put the relation of Earth’s gravity with Moon’s gravity and one will get the range relation between Earth and Moon.