a pillow , a textbook and a paper airplane are dropped from the top of a tall building at the same time. consider what you have
2 answers:
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Answer:
Explanation:
The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:
Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.
Also, we know that the centripetal force of an object describing a circular motion is given by:
Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.
Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:
Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So and
(Since
). Then, we get:
In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).
Answer:
Hello your question is incomplete below is the complete question
Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000
answer : V = 1.624* 10^-5 m/s
Explanation:
First we have to calculate the value of a
a = 93 * 10^6 mile/m * 1609.344 m
= 149.668 * 10^8 m
next we will express the distance between the earth and the sun
--------- (1)
a = 149.668 * 10^8
E (eccentricity ) = ( 1/60 )^2
= 90°
input the given values into equation 1 above
r = 149.626 * 10^9 m
next calculate the Earths velocity of approach towards the sun using this equation
------ (2)
Note :
Rc = 149.626 * 10^9 m
equation 2 becomes
(
therefore : V = 1.624* 10^-5 m/s