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:
Let \(W_e\) be the weight on Earth and \(W_m\) be the weight on the Moon.
Here, \(g_e\) is the acceleration due to gravity on Earth, and \(g_m\) is the acceleration due to gravity on the Moon.
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:
\( \frac{W_e}{W_m} = \frac{g_e}{g_m} \)
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.
The ratio of the weight of an object on the Earth to its weight on the Moon (\(W_e/W_m\)) is approximately 6.
In an elevator, the actual weight of a person is equal to the apparent weight when:
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.
How far below the Earth's level will the value of gravitational acceleration be half of the gravitational acceleration on the Earth's surface? (Radius of Earth 6400 km)
The radii of two planets are respectively R 1and R 2and their densities are respectively ρ 1and ρ 2. The ratio of the acceleration due to gravity (g 1/g 2) at their surface is