Assume a planet is a uniform sphere of radius R that (somehow) has a narrow radial tunnel through its center. Also assume we can position an apple anywhere along the tunnel or outside the sphere. Let FR be the magnitude of the gravitational force on the apple when it is located at the planet's surface. How far from the surface (in terms of R) is there a point where the magnitude of the gravitational force on the apple is 1/7FR if we move the apple (a) away from the planet and (b) into the tunnel?
I have no idea how to start this question please help!
To solve this question, we can consider Newton's Law of Universal Gravitation, which states that the gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Given that the planet is a uniform sphere, we can assume its mass is evenly distributed throughout its volume. Therefore, the force exerted on an object by the planet can be thought of as if all of the planet's mass were concentrated at its center.
Now let's break down the problem into two parts:
Part (a): Finding the distance from the surface where the gravitational force is 1/7FR when the apple is moved away from the planet.
In order to solve this, we need to use the formula for the gravitational force, F, which can be expressed as:
F = (G * (m1 * m2)) / r^2,
Where G is the gravitational constant, m1 is the mass of the planet, m2 is the mass of the apple, and r is the distance between the planet's center and the apple.
We are given that the gravitational force on the apple at the planet's surface is FR, so we can write this as:
FR = (G * (m1 * m2)) / R^2,
where R is the radius of the planet.
We are also given that the magnitude of the gravitational force on the apple at the desired point is 1/7FR.
Let's denote the distance from the surface to this point as d. Therefore, the distance from the apple to the planet's center would be R + d.
Now, let's set up the equation for the gravitational force at the desired point:
(1/7FR) = (G * (m1 * m2)) / (R + d)^2.
To find the value of d, we can rearrange the equation:
d = sqrt(((7 * G * (m1 * m2)) / FR) - R^2) - R.
This formula gives us the distance from the surface (in terms of R) where the magnitude of the gravitational force on the apple is 1/7FR when the apple is moved away from the planet.
Part (b): Finding the distance from the surface where the gravitational force is 1/7FR when the apple is moved into the tunnel.
In this case, when the apple is moved into the tunnel, the distance between the apple and the planet's center changes. Let's denote the distance from the center of the planet to the center of the tunnel as t.
Now, the distance from the apple to the planet's center is (R - t), and we need to find the value of t, where the gravitational force is 1/7FR.
Similar to part (a), we can set up the equation for the gravitational force at the desired point:
(1/7FR) = (G * (m1 * m2)) / (R - t)^2.
Rearranging the equation, we get:
t = R - sqrt(((7 * G * (m1 * m2)) / FR)).
This gives us the distance from the surface (in terms of R) where the magnitude of the gravitational force on the apple is 1/7FR when the apple is moved into the tunnel.
By using these formulas, you can calculate the distances d and t in terms of R. Remember to plug in the appropriate values for the gravitational constant (G), the mass of the planet (m1), the mass of the apple (m2), and the given gravitational force at the planet's surface (FR) to obtain the final answers.
To solve this question, let's break it down into smaller steps:
Step 1: Define the gravitational force on the apple at the planet's surface, FR.
The magnitude of the gravitational force on an object at the surface of a uniform sphere can be determined using Newton's law of universal gravitation:
F = G * (m1 * m2) / r^2
Where:
F is the gravitational force
G is the gravitational constant (approximately 6.67430 × 10^-11 N m^2/kg^2)
m1 and m2 are the masses of the two objects (in this case, the apple and the planet)
r is the distance between the centers of the two objects (in this case, the radius of the planet)
Since the mass of the apple doesn't affect the force acting on it, let's consider m1 = 1 kg and m2 = mass of the planet.
Step 2: Calculate the gravitational force on the apple at the planet's surface, FR.
Substituting the values into the formula, we have:
FR = G * (m1 * m2) / R^2
Step 3: Determine the distance from the surface where the magnitude of the gravitational force on the apple is 1/7FR when it is moved away from the planet.
Let the distance from the surface be denoted as x, where x is greater than R.
Step 4: Calculate the gravitational force on the apple when it is at a distance x away from the surface.
The distance between the apple and the center of the planet is R + x, and using the formula for the gravitational force, we have:
F = G * (m1 * m2) / (R + x)^2
Step 5: Set up and solve the equation to find x.
We want to find the value of x that makes the magnitude of the gravitational force 1/7 times FR:
1/7 * FR = G * (m1 * m2) / (R + x)^2
Substitute FR from Step 2:
1/7 * (G * (m1 * m2) / R^2) = G * (m1 * m2) / (R + x)^2
We can cancel out G, m1, and m2:
1/7 * (1 / R^2) = 1 / (R + x)^2
Step 6: Simplify and solve for x.
To simplify the equation, let's multiply both sides by R^2:
1/7 = (R + x)^-2
Take the reciprocal of both sides:
7 = (R + x)^2
Square root both sides:
√7 = R + x
Subtract R from both sides:
x = √7 - R
This is the distance from the surface where the magnitude of the gravitational force on the apple is 1/7FR when it is moved away from the planet.
Step 7: Determine the distance from the surface where the magnitude of the gravitational force on the apple is 1/7FR when it is moved into the tunnel.
When the apple is moved into the tunnel, it is at a distance of R from the center of the planet. Therefore, x = R.
This means that the distance from the surface where the magnitude of the gravitational force on the apple is 1/7FR is √7 - R when the apple is moved away from the planet and R when the apple is moved into the tunnel.
If the planet is uniform density, then the net gravity force is proportional to the volume enclosed by the radius.
1/7 (4/3 PI R^3)=4/3 PI r^3
so r/R= cuberoot 1/7
how far from the surface? H=R-r
H= R(1-cubroot 1/7) check that.