Give the other person brainest I was gonna answer but found no need if you get me
<h3><u>Question: </u></h3>
The equation for the speed of a satellite in a circular orbit around the Earth depends on mass. Which mass?
a. The mass of the sun
b. The mass of the satellite
c. The mass of the Earth
<h3><u>Answer:</u></h3>
The equation for the speed of a satellite orbiting in a circular path around the earth depends upon the mass of Earth.
Option c
<h3><u>
Explanation:
</u></h3>
Any particular body performing circular motion has a centripetal force in picture. In this case of a satellite revolving in a circular orbit around the earth, the necessary centripetal force is provided by the gravitational force between the satellite and earth. Hence
.
Gravitational force between Earth and Satellite: 
Centripetal force of Satellite :
Where G = Gravitational Constant
= Mass of Earth
= Mass of satellite
R= Radius of satellite’s circular orbit
V = Speed of satellite
Equating
, we get
Speed of Satellite 
Thus the speed of satellite depends only on the mass of Earth.
Hello!!
Here we have a simple matter of conservation of energy. ME=PE+KE.
At point A we have PE=mgh and KE=1/2mv^2. At point A all we have is PE since the coaster isn’t rolling yet. But by conservation of energy, we know that it will have enough energy to roll down and get to and equal height on another hill. Providing we are neglecting friction and drag and resistance forces which we are in this case. So we can conclude that the KE will be greater at Point B since ME=PE+KE and for ME to remain the same and we know the PE is less on lower hill, so we can conclude that KE on lower hill is greater to keep ME the same and have conservation of energy.
Hope this helps you understand the concept!! Any questions please just ask!! Thank you so much!!