Question

An Apple with the mass of 200g falls from the tree.what is the acceleration of the apple towards the earth.what is the acceleration of the earth towards the apple.(given mass of earth. and radius of earth and gravitational constant G=6.67*10^-11​

Answer 1

Answer:

Assume that the earth is a sphere (with radius equal to its actual radius at the equator) of uniform density.

• Acceleration of the apple towards the earth: approximately $9.79\; \rm m \cdot s^{-2}$.

• Acceleration of the earth towards the apple: approximately $3.28 \times 10^{-25}\; \rm m \cdot s^{-2}$.

Both values were rounded to three significant figures. Air resistance is assumed to be negligible.  

Explanation:

Convert the mass of the apple to standard units:

$m(\text{apple}) = 200\; \rm g = 0.200\; \rm kg$.

Look up the mass and radius of the earth:

• Mass of the earth: $m(\text{earth})\approx 5.97\times 10^{24}\; \rm kg$.
• Radius of the earth: $r \approx 6.3781\times 10^{6}\; \rm m$ (at the equator.)

Assume that the earth is a sphere of uniform density. The acceleration of an object moving towards the earth under free fall would be approximately equal to the gravitational field strength of the earth at that position.

For a sphere of mass $m$ (assuming uniform density,) the gravitational field strength $g$ at a distance of $r$ away from the center of the sphere would be:

$\displaystyle g = \frac{G \cdot m}{r^2}$.

The height of the tree should be much smaller than the radius of the earth. Therefore, the distance between the apple and the center of the earth would be approximately equal to the radius of the earth.

The strength of the gravitational field of the earth at the position of the apple would be approximately:

$\begin{aligned} & \frac{G \cdot m(\text{earth})}{r^2} \\ &\approx \frac{6.67\times 10^{-11}\; \rm m^3\cdot kg^{-1}\cdot s^{-2} \times 5.97 \times 10^{24}\; \rm kg}{\left(6.3781\times 10^{6}\; \rm m\right)^2} \\ &\approx 9.79\; \rm m\cdot s^{-2} \end{aligned}$.

This value should be approximately equal to the acceleration of the apple towards the earth.

On the other hand, assume that the apple acts like a point mass when compared to the earth. In other words, assume that the radius of the apple is much smaller than the distance between the apple and the earth.

It can be shown (using calculus) that if the earth is a sphere with uniform density, the earth will behave just like a point mass at the center of the earth when studying interactions between the earth and objects outside of it.

At the center of the earth, the strength of the gravitational field due to the apple would be:

$\begin{aligned} & \frac{G \cdot m(\text{apple})}{r^2} \\ &\approx \frac{6.67\times 10^{-11}\; \rm m^3\cdot kg^{-1}\cdot s^{-2} \times 0.200\; \rm kg}{\left(6.3781\times 10^{6}\; \rm m\right)^2} \\ &\approx 3.28 \times 10^{-25}\; \rm m\cdot s^{-2} \end{aligned}$.

Under the assumption that the earth is a sphere of uniform density, this value should be equal to the acceleration of the earth towards the apple due to the gravitational attraction of the apple.