Calculate the force of gravity on a spacecraft

AI Thread Summary
To calculate the force of gravity on a spacecraft 19,200 km above Earth's surface with a mass of 1500 kg, the gravitational force equation Fg=[(G)(M spacecraft)(Mearth)]/r^2 is used. The distance r must be measured from the center of the Earth, which means it is 4 Earth radii (the altitude plus Earth's radius). The correct calculation yields a gravitational force of approximately 915.98 N after converting units properly. It's crucial to read exam questions carefully to avoid common pitfalls related to altitude versus radius. Understanding the distinction between these measurements is essential for accurate calculations in gravitational force problems.
mathcrzy
Messages
47
Reaction score
0
1. Homework Statement

Calculate the force of gravity on a spacecraft 19200 km (3 Earth radii) above the Earth's surface if its mass is 1500 kg.


2. Homework Equations

Fg=[(G)(M spacecraft )(Mearth)]/r^2=ma.

3. The Attempt at a Solution

a=[(6.67e-11)(1500)(6e24)]/19200000^2(convert km to m)=1628.4
 
Physics news on Phys.org
mathcrzy said:
1. Homework Statement

Calculate the force of gravity on a spacecraft 19200 km (3 Earth radii) above the Earth's surface if its mass is 1500 kg.


2. Homework Equations

Fg=[(G)(M spacecraft )(Mearth)]/r^2=ma.

3. The Attempt at a Solution

a=[(6.67e-11)(1500)(6e24)]/19200000^2(convert km to m)=1628.4

Whoops! Ya fell into the pit -- they catch lots of students this way (especially on multiple choice questions). What does r represent in Newton's gravitational force equation? The distance you are given is measured from the surface of the Earth, that is, it is an altitude... (Oh, and don't forget to put units on your answers!)
 
r represents the radius and so it would be 4 Earth radii because they only talk about from the surface? so if that is the case i would do a=[(6.67e-11)(1500)(6e24)]/25600000^2(convert km to m)=915.98N
 
mathcrzy said:
r represents the radius and so it would be 4 Earth radii because they only talk about from the surface? so if that is the case i would do a=[(6.67e-11)(1500)(6e24)]/25600000^2(convert km to m)=915.98N

That looks fine! Be sure to read exam problems carefully because little traps like this are frequently set for the unwary. (Altitude is more convenient for describing orbits, and people who work with Earth satellites do so all the time, but the value always has to be adjusted to distance from the center of Earth for use in equations.)
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top