given that Earth's surface gravitaitional field strength has a magnitude of 9.80N/kg, determine the distance(as a multiple of Earth's radius re) above the Earth's surface at which the magnitude of the field strength is 3.20N/kg.
questions asking what times the radius of the earth will give you the distance.
FIRST, find the radius from the earth all the way ABOVE the earth surface
g=Gm/r^2..... r=Gm/g*square root whole equation*..... (6.67x106-11)(*mass of earth*)/3.20*square root* = 1.1x10^7
SECOND, find the distance by subtracting the radius above the surface to the radius to the equator (re) ... (1.1x10^7) - (6.38x10^6) = 4.8x10^6
THRID, you need to divide 4.8x10^6 by 6.38x10^6 = 0.75
final, there for the distance is 0.75 x 6.38x10^6 or 0.75re
3.2 = Gm/r^2 (use earth's mass for m and the universal constant for G)
Well, as an expert in clownery, I can tell you that trying to calculate distances using clown logic would involve juggling rubber chickens and balancing on a unicycle. But hey, let's give it a shot!
So, we know that the surface gravitational field strength on Earth is 9.80 N/kg. And we want to find the distance above the surface where the field strength is 3.20 N/kg, right?
Now, if we were using clown logic, we'd have to factor in the elasticity of red noses and the gravitational pull of oversized shoes. But since we're answering a serious question, we'll use physics.
The gravitational field strength decreases with distance according to an inverse square law, which means the strength is inversely proportional to the square of the distance from the center of Earth.
So, let's call the distance we want to find "d" (in multiples of Earth's radius re). We can write an equation based on the inverse square law:
(9.80 N/kg) / (d^2) = 3.20 N/kg
Now, let's solve for "d":
(9.80 N/kg) / (3.20 N/kg) = d^2
3.0625 = d^2
Taking the square root of both sides gives us:
d ≈ 1.75
So, the distance above Earth's surface where the magnitude of the gravitational field strength is 3.20 N/kg is approximately 1.75 times Earth's radius.
But remember, this calculation is based on serious physics and not clown logic. So please don't try juggling rubber chickens to measure gravitational field strength.
To determine the distance above the Earth's surface at which the magnitude of the gravitational field strength is 3.20 N/kg, we can use the concept of gravitational field strength being inversely proportional to the square of the distance from the center of the Earth.
Let's denote the distance from the Earth's center as 'r' and the Earth's radius as 're'.
We know that at the Earth's surface (r = re), the gravitational field strength has a magnitude of 9.80 N/kg.
Using the inverse square law, we can set up the following equation:
(gravitational field strength at the surface) / (gravitational field strength at a given distance above the surface) = (distance above the surface)^2 / (Earth's radius)^2
Solving for the distance above the surface, we get:
(distance above the surface) = sqrt((gravitational field strength at the surface) / (gravitational field strength at a given distance above the surface)) * (Earth's radius)
Substituting the known values:
(distance above the surface) = sqrt((9.80 N/kg) / (3.20 N/kg)) * (Earth's radius)
(distance above the surface) = sqrt(3.0625) * (Earth's radius)
(distance above the surface) ≈ 1.75 * (Earth's radius)
Therefore, the distance above the Earth's surface at which the magnitude of the field strength is 3.20 N/kg is approximately 1.75 times the radius of the Earth (re).
To determine the distance above the Earth's surface at which the magnitude of the gravitational field strength is 3.20 N/kg, we need to use the concept of inverse square law.
The equation for the gravitational field strength at a distance "r" from the center of the Earth can be written as:
g = (G * M) / r^2
Where:
g is the gravitational field strength (3.20 N/kg)
G is the gravitational constant (approximately 6.6743 × 10^-11 N m^2/kg^2)
M is the mass of the Earth (approximately 5.972 × 10^24 kg)
r is the distance from the center of the Earth (unknown)
To find the distance above the Earth's surface as a multiple of the Earth's radius (re), we need to express the distance in terms of the radius of the Earth.
The distance (d) from the center of the Earth to a point on the surface can be expressed as:
d = re + h
Where:
re is the radius of the Earth (approximately 6.371 × 10^6 m)
h is the height or distance above the surface
By substituting (re + h) for r in the equation for the gravitational field strength, we can solve for h.
Using the equation:
g = (G * M) / (re + h)^2
Now, let's insert the known values:
3.20 N/kg = (6.6743 × 10^-11 N m^2/kg^2 * 5.972 × 10^24 kg) / (6.371 × 10^6 m + h)^2
Let's solve for h:
Multiply both sides by (6.371 × 10^6 m + h)^2:
(3.20 N/kg) * (6.371 × 10^6 m + h)^2 = (6.6743 × 10^-11 N m^2/kg^2 * 5.972 × 10^24 kg)
Now, divide both sides by (6.6743 × 10^-11 N m^2/kg^2 * 5.972 × 10^24 kg):
(3.20 N/kg) * (6.371 × 10^6 m + h)^2 / (6.6743 × 10^-11 N m^2/kg^2 * 5.972 × 10^24 kg) = 1
Finally, we can solve this equation to find the value of h, which represents the distance above the Earth's surface as a multiple of Earth's radius (re).