If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be 
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<h3>What is an escape velocity?</h3>
The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.
The formula to calculate the escape velocity of earth is given below:-
Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-
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Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be .
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Answer:
6.49 x 10^-8 N
Explanation:
formula is
F= G * ((m1 * m2)/r^2)
F = 6.67x10^-11 * ((6.8*6.8/.218)
F = 6.49 x 10^-8 Newtons
Answer:
what did u say and what language are you speaking in
Hi there!
Recall the equation for centripetal force:
We can rearrange the equation to solve for 'r'.
Multiply both sides by r:
Divide both sides by Fc:
The initial height of the first body is given by:
where
g is the gravitational acceleration
t is the time it takes for the body to reach the ground
Substituting t=1 s, we find
The second body takes takes t=2 s to reach the ground, so it was located at an initial height of
The second body started its fall 1 second before the first body, therefore when the second body started its fall, the first body was located at its initial height, i.e. at 4.9 m from the ground.
