Need help with High School Physics PLEASE

[starvinmarvin]

Hey everyone. I’m in high school Physics and I need some help. We are in a unit involving gravity, mass, and radius and heights of Earth. Here are some equations and givens we have to help:

M1g= (Gm1m2)/r^2
G= 6.67 x 10^-11
R= 6.38 x 10^6
V= Square root of (Gmp/r)
T= Square root of (4pi^2r^3/Gmp)

Any help would be much included. I have tried working on these problems for like 45 minutes already and I can't come up with anything because I don't really even know how to start or continue on these! Please try to include work with your help. THANK YOU!

#7: Calculate the period of Earth’s rotation and its velocity for people at the equator to feel weightless. (Hint: what would FN be?) How long would an Earth day be then? What type of adjustments would we have to make to live in such a situation?

MORE INFO FROM TEACHER:
-Fg= -Fc
-mpg= (4pi^2mpr)/T^2 on eartch

#9: At what height above Earth’s surface is gravity half of what it is on the Earth’s surface?

#11: What is the gravitational field strength of a 5.00N object at a distance h above Earth with a mass of 1.34kg? What is the weight of another object at this same height if its mass is 188kg? What is the weight on each kg of mass at this height? What is the gravitational acceleration at this height? What is a force field? Draw the relative gravitational field strength for both objects at h and at a distance of 2h and 1/2h with arrows.

#13: Calculate the height above Mercury that an astronaut would have to be so that her weight would be half that on Mercury? (Hint: r= 2.44 x 10^6)

 

[YellowTaxi]

Hey everyone. I’m in high school Physics ...

#9: At what height above Earth’s surface is gravity half of what it is on the Earth’s surface?

for #9 all you need is: F = GMm/r^2

If you want F to become half then you need to make the number r^2 go to twice it's original value. (so that all of GMm/r^2 will go to half its original value)

So you would have to make r change to 1.4142 times the initial value [ 1.4142 is the square root of 2]

On Earth's surface r= 6.38 x 10^6m
so you just need to change that to 1.4142 x 6.38 x 10^6m
F will drop to half.

sorry I didn't understand the facts & figures that you gave, but if you can see how #9 is so easy to solve, that alone might help you work out the other problems as well.

 

[Moetasim]

Hi there! It sounds like you are working on some challenging concepts in your high school Physics class. Don't worry, with some practice and understanding of the equations and givens provided, you will be able to solve these problems. Let's break down each question and provide some guidance:

#7: To calculate the period of Earth's rotation and its velocity for people at the equator to feel weightless, we need to use the equation T = √(4π^2r^3/Gmp). Here, T represents the period (or time) of Earth's rotation, r is the radius of Earth (given as 6.38 x 10^6), G is the gravitational constant (given as 6.67 x 10^-11), m is the mass of Earth (given as 5.98 x 10^24), and p represents the density of Earth (which we can assume to be around 5.52 x 10^3 kg/m^3).

To find the velocity, we can use the equation V = √(Gmp/r). Substituting the given values, we get V = √(6.67 x 10^-11 x 5.98 x 10^24 / 6.38 x 10^6) = 7.91 x 10^3 m/s. This means that at the equator, the velocity needed for people to feel weightless is approximately 7.91 km/s.

The Earth's rotation period can be calculated by dividing the circumference of Earth (2πr) by the velocity (V). This gives us a rotation period of 5066 seconds or approximately 84 minutes. This means that an Earth day would be 84 minutes long, which is significantly shorter than our current 24-hour day.

To live in such a situation, we would need to make adjustments to our daily routines, such as sleeping, eating, and working, to accommodate the shorter day. It would also affect agriculture, animal behavior, and many other aspects of life on Earth.

#9: To calculate the height above Earth's surface where gravity is half of what it is on the surface, we can use the equation M1g = (Gm1m2)/r^2. Here, M1 represents the mass of the object experiencing gravity (which we can assume to be 1 kg), g is the gravitational field strength (which