Планета Марс, масса которой равна0,11 массы Земли, удалена от...

à 24 avril

v=sqrt(GM/(R+3R)=sqrt(6.67*10^-11*6*10^24(масса Земли)/4*6.37*10^6(радиус Земли)=sqrt(15.7*10^6)=3.96*10^3м/c = 3.96км/c.we need to find the mass of the planet. Vo = sqrt ( GM/r).The force of gravity at a given radius from the center of earth, according to Newton's law of gravitation, is g=(GM)/(r^2), where G is the universal gravitation constant, and M is the mass of the parent body. v = sqrt(GM/6738000). G is given as 6.67384x10-11 Nm^2/kg^2.Б). $ mv^2/R=G\frac{mM}{R^2} \\ v=\sqrt{GM/R} \\ v_g / v_m= \frac{\sqrt{GM/R}}{\sqrt{GM/1.52R}}=1/ \sqrt{1.52}=1.23 \\ v_m=v_g/1.23=24.4 m/s $.

What is the orbital speed of this second satellite? | Yahoo Answers

What I would like to know is why eqn $(3)$ is not a valid escape velocity equation? Or, put in another way, mathematically, the derivation in $(2)$ seems sound; yet it is out by a factor of $\sqrt{2}$. What is 'missing' from the derivation $(2)$? EDIT: As I mentioned in the comment below, just to be clear, I...G*M*m/R^2 = m*v^2/R G*M/R = v^2 v = sqrt(GM/R) = sqrt(GM/R0^2 * R0^2/R) = sqrt(g * R0^2/R) T = 2*pi*R/v = 2*pi/sqrt(g * R0^2/R^3) = 2*3.14/sqrt(9.81*6371000^2/24520000^3) = 38200 сек = 10 часов 37 мин.v = sqrt((GM)/(R+h)).

$What is the orbital speed of this second satellite? | Yahoo Answers$

Orbital Calculations

Если да то v=sqrt(GM/r). Ответить. Привет!Cancelling the m's and a factor of r gives: GM/r = v2. We can bring in the period using: v = 2πr / T. This gives GM/r = 4π2r2/T2.Re: Using v = sqrt (GM/r)? « Reply #1 on: June 16, 2013, 10:27:20 pm ». But I'm reading your explanation and I understand everything except 'For the satisfying of the magnitudes: GM/r^2=v^2/r' So when you say 'satisfying of the magnitudes' what do you actually mean by that?v = sqrt((GM)/(R+h)).

Think about moving your arms in a circle. You know that if you push harder on your arms, one of three things will happen:1, increase the speed of rotation2, decrease the radius of the circle3, a combination of the above

Also, changing the speed seems to be so much harder than decreasing the radius of the circle. This is because the force mv^2/r, directed radially inward, is a necessary prerequisite for the object to move in a circular path of radius r and velocity v (uniform circular motion just means no acceleration in the direction of the object's movement; that's not needed for this year). If you ever slack off with your arms, either you have to increase the radius or the speed drops. The force mv^2/r is not a force you get; it's what you need.

A circular orbit is an example. If a satellite does not quite have the force to move in that orbit, it will move out further until the gravitational force it receives matches the centripetal acceleration. If you could magically slow the satellite down, that would work too. Don't worry about the possibility of elliptical orbits in VCE. However, if the satellite receives a net force which is directed radially inward, of magnitude mv^2/r and is perpendicular to its motion, it will move in that circle. Hence why I can equate GMm/r^2, the gravitational force, with mv^2/r, for the requirement of uniform circular motion.