Calculate the force of Earth's gravity on a spacecraft

AI Thread Summary
The discussion revolves around discrepancies between the user's calculations and the answers provided by the mastering physics program for three problems related to gravitational force and satellite velocities. For the first problem, the user calculates the gravitational force on a spacecraft at 2.5 Earth radii but gets a different answer than the program. In the second problem, the user determines the speeds of two satellites at different altitudes and finds a factor of 1.74, while the program states 1.4. The third problem involves calculating the altitude where gravity is half that at the surface, with the user arriving at approximately 9,022,262 m, contrasting with the program's 2,640,000 m. The user expresses confusion over the correct interpretation of the radius in relation to altitude, questioning whether the homework program or their calculations are flawed.
cabrady92
Messages
2
Reaction score
0
I am in a classical physics class and the class requires that we use mastering physics for our homework. The only problem is that I can't get what mastering physics claims is the right answer to these problems. Can someone try these calculations to see if I am right or wrong?


Problem 6.1 : Calculate the force of Earth's gravity on a spacecraft 2.50 Earth radii above the Earth's surface if its mass is 1400 kg.
My calculations: F= G((m1mE)/r^2) = (6.67*10^-11)*((1400kg)*(5.97*10^24kg))/(6378km*2.5*10^3)^2= 2192.7 N

Their answer: 1120 n?


Problem 6.28: Two satellites orbit Earth at altitudes of 5600 km and 1.7×104 km.Which satellite is faster? By what factor? = vclose/vfar.
My calculations: vclose = sqrt(G(mE/r)), sqrt((6.67*10^-11)*(5.98*10^24kg)/(5600km*10^3)) = 8439.55 m/s

vfar = sqrt(G(mE/r)), sqrt((6.67*10^-11)*(5.98*10^24kg)/(17400*10^3)) = 4843.83 m/s

vclose/vfar = 1.74

Their answer: 1.4?

Problem 6.51: How far above the Earth's surface will the acceleration of gravity be half what it is at the surface?

My calculations: m1(1/2g) = Gm1mE/r^2, r^2 = 2GmE/g, r = sqrt(2GmE/g), r = sqrt( 2*(6.67*10^-11)*(5.98*10^24)/9.8) = 9022262 m

Their answer: 2640000 m?


Can you try solving these answers and tell me what I am doing wrong or if the homework program is flawed?
 
Physics news on Phys.org
I'm even more confused... the book tells me to use r as altitude but mastering phyics tells me that r is the altitude + r...
 
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top