Asked by Bobbie
What is the gravitational field strength at a place 220km above Earth's surface, the altitude of many piloted space flights?
Answers
Answered by
bobpursley
Newtons law of gravity:
g=GMe/(radiusEarth + altitude)^2
g=GMe/(radiusEarth + altitude)^2
Answered by
Bobbie
k well i don't know what Me stands for or what the altitude is or how to find it.
Answered by
bobpursley
You should know Newtons Law.
Me is mass of the Earth.
altitude is 220Km, given.
Me is mass of the Earth.
altitude is 220Km, given.
Answered by
Bobbie
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
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
Answered by
Bobbie
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
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
Answered by
chloe
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)
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)
Answered by
Heather
@Bobbie
0.22m should be 220000m
The original question gave 220km (kilometres), which converts into 220000m (metres). Just a simple conversion mistake, great job!
0.22m should be 220000m
The original question gave 220km (kilometres), which converts into 220000m (metres). Just a simple conversion mistake, great job!
Answered by
Evan
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 ?
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 ?
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