What is the gravitational field strength at a place 220km above Earth's surface, the altitude of many piloted space flights?
@Bobbie
0.22m should be 220000m
The original question gave 220km (kilometres), which converts into 220000m (metres). Just a simple conversion mistake, great job!
Newtons law of gravity:
g=GMe/(radiusEarth + altitude)^2
k well i don't know what Me stands for or what the altitude is or how to find it.
You should know Newtons Law.
Me is mass of the Earth.
altitude is 220Km, given.
well my teacher cant teach so i dont. and i used the formula but i got a different answer then the back of the book. i don't get what im doing wrong.
g=GMe/(radius Earth+altitude)^2
g=(6.67x10*-11Nm^2/kg^2)x(5.98x10*24kg)/ (6.38X10*6m+0.22m)^2
g= (3.99x10*14Nm^2/kg)/(4.07x10*13m^2)
g=9.8N/kg
the back of the book says 9.1N/kg
well my teacher cant teach so i dont. and i used the formula but i got a different answer then the back of the book. i don't get what im doing wrong.
g=GMe/(radius Earth+altitude)^2
g=(6.67x10*-11Nm^2/kg^2)x(5.98x10*24kg)/ (6.38X10*6m+0.22m)^2
g= (3.99x10*14Nm^2/kg)/(4.07x10*13m^2)
g=9.8N/kg
the back of the book says 9.1N/kg
A simple pendulum is set up on a spacecraft. The spacecraft hovers at an altitude of
250 km above the surface of Mars. The length of the pendulum is 20 cm, and the mass
of the bob is 0.2 kg.
a) What is the strength of the gravitational field �, due to Mars, as measured in the
spacecraft? (5 pts.)
b) What is the period of the pendulum? (5 pts)
c) Suppose the tension at the very bottom of the trajectory is 3 N. What is the speed of
the pendulum mass at that location? (5 pts)
d) Now imagine that the string snaps at the bottom of the trajectory, and the mass moves
off at a tangent. If the lowest point of the pendulum’s path is 1.5 m above the floor of
the spaceship, what horizontal distance does the mass travel (10 pts)
How did you get the formula g=GMe/(radiusEarth + altitude)^2
my prof. gave me a different formula without the "+altitude" so the formula i have is: g=GMe/radiusEarth^2
so how did you get the +altitude ?
To find the gravitational field strength at a specific altitude above the Earth's surface, we can use Newton's law of universal gravitation. According to this law, the gravitational force between two objects is given by the equation:
F = (G * m1 * m2) / r^2
Where:
F is the gravitational force between the two objects,
G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2),
m1 and m2 are the masses of the two objects, and
r is the distance between the centers of the two objects.
In this case, the two objects are the Earth and an object at an altitude of 220 km above the Earth's surface. To simplify the calculation, we can assume that the object's mass is negligible compared to the mass of the Earth.
Now, let's calculate the gravitational field strength at this altitude:
Step 1: Convert the altitude from kilometers to meters.
220 km = 220,000 meters
Step 2: Determine the distance between the object and the center of the Earth.
The radius of the Earth is approximately 6,371 km (or 6,371,000 meters).
So, the distance between the object and the center of the Earth is:
distance = radius + altitude
distance = 6,371,000 + 220,000
distance = 6,591,000 meters
Step 3: Calculate the gravitational field strength using the equation:
g = (G * M) / r^2
Where:
g is the gravitational field strength,
G is the gravitational constant (6.67430 × 10^-11 m^3 kg^-1 s^-2), and
M is the mass of the Earth (approximately 5.972 × 10^24 kg).
g = (6.67430 × 10^-11 * 5.972 × 10^24) / (6,591,000)^2
After evaluating this expression, the gravitational field strength at a place 220 km above Earth's surface is approximately 7.25 m/s^2.