When you drop a 0.36 kg apple, Earth exerts a force on it that accelerates it at 9.8 m/s^2 toward the earth's surface. According to Newton's third law, the apple must exert an equal but opposite force on Earth. If the mass of the earth 5.98x10^24 kg, what is the magnitude of the earth's acceleration toward the apple? (the answer must be in units of m/s^2)
I tried this and used F=ma
(.36)(9.8) = (5.98x10^24)(a)
so a= 5.899x10^-25
But my answer was incorrect. Can someone tell me how to solve this problem and what I did wrong?
Well, it seems like you're already on the right track by using Newton's second law, F = ma. However, it looks like you made a small mistake when setting up the equation.
The force exerted by the apple on Earth is equal in magnitude but opposite in direction to the force exerted by Earth on the apple. Therefore, the equation should be:
(.36)(9.8) = (5.98x10^24)(-a)
Since the two forces have opposite directions, we need to include a negative sign when solving for the acceleration of Earth. Now let's solve for a:
a = (-(.36)(9.8)) / (5.98x10^24)
a ≈ -5.858x10^-26 m/s^2
Therefore, the magnitude of Earth's acceleration towards the apple is approximately 5.858x10^-26 m/s^2. Keep in mind that the negative sign simply indicates the opposite direction of acceleration compared to the apple.
To solve this problem, you need to apply Newton's law of universal gravitation. It states that the force of attraction between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Mathematically, it can be expressed as:
F = G * (m1 * m2) / r^2
Where:
- F is the force of attraction
- G is the gravitational constant (approximately 6.67 × 10^-11 N m^2 / kg^2)
- m1 and m2 are the masses of the objects
- r is the distance between the centers of the objects
In this case, the apple and the Earth create a gravitational force of attraction that causes them to accelerate towards each other. The force exerted on the apple by the Earth is given by:
F = m1 * a
Where:
- F is the force exerted by the Earth on the apple
- m1 is the mass of the apple
- a is the acceleration due to gravity, which is 9.8 m/s^2
According to Newton's third law, the apple exerts an equal but opposite force on Earth. This means that the force exerted by the apple on Earth is also F.
So, you can equate the two forces:
m1 * a = F = G * (m1 * m2) / r^2
Now, solve for the acceleration of Earth (a2), considering that m1 is the mass of the apple (0.36 kg), m2 is the mass of the Earth (5.98 × 10^24 kg), and the distance between the apple and Earth (r) is negligible as they are in contact:
a2 = G * m1 / r^2
Plug in the given values:
a2 = (6.67 × 10^-11 N m^2 / kg^2) * (0.36 kg) / (0.0 m)^2
Simplifying, you get:
a2 = 0 (approximately)
Therefore, the magnitude of the Earth's acceleration towards the apple, considering the mass of the Earth and the given conditions, is approximately 0 m/s^2.