The satellite is 8.02 × 10⁵ m above Earth's surface.
Let H be the height above the surface of the Earth; since we know that the satellite is rotating around the Earth due to the gravitational pull of the planet, we may assert
Procedure to solve:
F = mv²/R+H
H = mv²/F - R
H = (1160 × 7446²/8955 - 6.38 × 10⁶)
M = 8.02 × 10⁵ m
About centripetal force:
The force applied to an item that is in velocity of curved motion that is pointed toward the axis of rotation or the centre of curvature is known as a centripetal force.
The centripetal force formula is given as the product of mass (in kg) and tangential velocity (in meters per second) squared, divided by the radius (in meters) that implies that on doubling the tangential velocity, the centripetal force will be quadrupled. Mathematically it is written as:
F = mv²/r
Learn more about velocity here:
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Change in velocity of larger moose: (1/3)v - v = -(2/3)v
<span>change in velocity of small moose: (1/3)v - (-v) = (4/3)v </span>
<span>- (change in velocity of larger moose)/(change in velocity of smaller moose) = 2
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Answer:
The gravity on this planet is stronger than that of earth.
Explanation:
First we need to find the acceleration due to gravity value of this planet to compare its gravity force with that of the earth. Hence, we will use second equation of motion:
h = Vi t + (0.5)gt²
where,
h = height or depth of crater = 100 m
Vi = Initial Velocity of rock = 0 m/s
t = time = 4 s
g = acceleration due to gravity on this planet = ?
Therefore,
100 m = (0 m/s)(4 s) + (0.5)(g)(4 s)²
g = (200 m)/(16 s²)
g = 12.5 m/s²
on earth:
ge = 9.8 m/s²
Since,
ge < g
Therefore,
<u>The gravity on this planet is stronger than that of earth.</u>
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