Weight Ratio: Earth vs Moon Explained
The weight of an object depends on its mass and the acceleration due to gravity at its location. The formula for weight (\(W\)) is:
\(W = m \times g\)
Where:
- \(m\) is the mass of the object (constant).
- \(g\) is the acceleration due to gravity.
Calculating Weight on Earth and Moon
Let \(W_e\) be the weight on Earth and \(W_m\) be the weight on the Moon.
- Weight on Earth: \(W_e = m \times g_e\)
- Weight on Moon: \(W_m = m \times g_m\)
Here, \(g_e\) is the acceleration due to gravity on Earth, and \(g_m\) is the acceleration due to gravity on the Moon.
Determining the We/Wm Ratio
To find the ratio \(W_e / W_m\), we divide the weight on Earth by the weight on the Moon:
\( \frac{W_e}{W_m} = \frac{m \times g_e}{m \times g_m} \)
Since the mass (\(m\)) is constant, it cancels out:
\( \frac{W_e}{W_m} = \frac{g_e}{g_m} \)
Gravity Values and Ratio Calculation
The acceleration due to gravity on Earth (\(g_e\)) is approximately \(9.8 \, \text{m/s}^2\). The acceleration due to gravity on the Moon (\(g_m\)) is approximately \(1.62 \, \text{m/s}^2\).
Now, calculate the ratio:
\( \frac{W_e}{W_m} \approx \frac{9.8 \, \text{m/s}^2}{1.62 \, \text{m/s}^2} \approx 6.05 \)
Therefore, the ratio \(W_e / W_m\) is approximately 6.
Conclusion
The ratio of the weight of an object on the Earth to its weight on the Moon (\(W_e/W_m\)) is approximately 6.
Important Questions from Acceleration due to gravity of the earth
- If a ball is thrown vertically upwards with the speed $u$, the distance covered during the second-to-last $t$ seconds of its ascent is
(Assume $t$ is less than half of the total ascent time). A body freely falling from rest has acquired a velocity ‘v’ after it falls through a distance ‘h’. The distance it has to fall down further for its velocity to become double is:
On earth, the value of G = 6.67 × 10 -11 Nm 2kg -2 . What is the value on moon, where acceleration due to gravity is nearly one - sixth than that of earth?
What is the force required to produce an acceleration of 9.8 m/s 2on a body of weight 9.8N? Take g = 9.8 m/s 2.
At what height above the surface of the earth does the weight of an object reduce by 1%. Given the radius of the earth is 6400.
