Since we know that Mount Everest is 8848 meter high, we just have to add that to the radius of the Earth and find the radius of the centripetal force. We have that the radius is 6408848 meters.
Now, we need the formulas:
F=(mv^2)/r and
F=(GMm)/r^2
We combine the tow formulas and get
(mv^2)/r=(GMm)/r^2
Solve for v and get
v=sqrt(MG/r),
G being 6.67 x 10^-11
M being the mass of the Earth, or 6.4 x 10^24
and r we already solved
So, plug in, and the answer we get is 7902.2 m/s
Velocity of satellite and hence its linear momentum changes continuously due to the change in the direction of motion in a circular orbit. However, angular momentum is conserved as no external torque acts on the satellite.
Satellites orbit the Earth or other bodies due to a careful balance of their velocity and the gravitational attraction of the body. Essentially gravity pulls them down but their velocity moves then out (Newton's Fist Law of Motion) at the same rate. They keep missing the body they orbit.The path is not necessarily circular since the gravity over the Earth varies with the density of the ground below the satellite. They are also satisfied to be in an elliptical orbit (closer at some times than others). The moon is a good example of a satellite in an almost circular elliptical orbit. comets have wildly elliptical orbits.
You cannot. A distance vs time graph only measures radial distance - that is, distance from the origin to the object. If the object is going around the origin along a circular path, the distance vs time graph will not show any change in distance.The [incorrect] answer that you are required to give is that the graph will be a horizontal line during that period. But as explained above, the horizontal graph only means the object has no movement towards or away from the origin, not that it has no movement.
Its velocity.
say shell velocity = 1000 m/s , launch angle = 45 deg from horizontal , then horizontal component = cos 45 * 1000 = 0.707 *1000 = 707 m/s
circular velocity
If the velocity of the satellite is always perpendicular to the force of gravity, then the eccentricity of the orbit is zero, and it's perfectly circular.
circular.
the velocity will be velocity divided by square root of 2
Yes, the satellite is accelerating because it is revolving around our earth and in a circular motion so its velocity changes every second so it is accelerating.A2. No, the satellite is not accelerating. Acceleration is defined as the rate of change of velocity. But, its velocity is constant. The centrifugal effect is exactly balanced by the pull of gravity (assuming a circular orbit).But a nice question. The net acceleration between these forces is zero.
A satellite small enough to be treated as a point particle. Can earth's gravity exert a torque on a satellite about the earth's center? Torque causes an object to rotate around a specific point. Torque = force * perpendicular distance and Torque = moment of Inertia * angular acceleration. When a satellite is launched, it is forced up to a specific distance from the earth's center and accelerated to a specific velocity parallel to the surface of the earth. The satellite continues moving in circular orbit. The force which causes the satellite to move in a circular path is the gravitational force caused by the mass of the earth, mass of the satellite, and distance from the center of mass of the earth to the center of mass of the satellite. This force causes the direction of the velocity to rotate so it is always tangent to the circle. This force produces the torque which makes causes the satellite to rotate so the direction of its velocity is always perpendicular to the direction of the gravitational force.
initial velocity, angle of launch, height above ground When a projectile is launched you can calculate how far it travels horizontally if you know the height above ground it was launched from, initial velocity and the angle it was launched at. 1) Determine how long it will be in the air based on how far it has to fall (this is why you need the height above ground). 2) Use your initial velocity to determine the horizontal component of velocity 3) distance travelled horizontally = time in air (part 1) x horizontal velocity (part 2)
initial velocity, angle of launch, height above ground When a projectile is launched you can calculate how far it travels horizontally if you know the height above ground it was launched from, initial velocity and the angle it was launched at. 1) Determine how long it will be in the air based on how far it has to fall (this is why you need the height above ground). 2) Use your initial velocity to determine the horizontal component of velocity 3) distance travelled horizontally = time in air (part 1) x horizontal velocity (part 2)
If space were entirely empty this would be true, but even minute gravitational forces can change the trajectory and velocity of a projectile.
Velocity of satellite and hence its linear momentum changes continuously due to the change in the direction of motion in a circular orbit. However, angular momentum is conserved as no external torque acts on the satellite.
No, horizontal velocity and vertical velocity are independent and have no effect on each other.
Satellites orbit the Earth or other bodies due to a careful balance of their velocity and the gravitational attraction of the body. Essentially gravity pulls them down but their velocity moves then out (Newton's Fist Law of Motion) at the same rate. They keep missing the body they orbit.The path is not necessarily circular since the gravity over the Earth varies with the density of the ground below the satellite. They are also satisfied to be in an elliptical orbit (closer at some times than others). The moon is a good example of a satellite in an almost circular elliptical orbit. comets have wildly elliptical orbits.