2) what is the free fall acceleration? 3) what is min downward force? 7) Astrono

Question

2) what is the free fall acceleration?
3) what is min downward force? 7) Astronomers discover planet X in our solar system. Planet X is 10 times the distance from the sun than the Earth is. Planet X has a radius 4 times as big as th radius of the Earth, but only 1/2 of its mass. How much weaker or stronger is the gravitational force by the Sun on Planet X th it is on the Earth? [3 pts] a) b) What is the free-fall acceleration of Planet X near the surface compared to g on Earth? 13 pts] 8) What is the minimum downward force F slipping? Let _ 0.5 and : 0.2? 14pts) one must apply to keep the block from 200 N

Explanation / Answer

let mass of earth=M

radius of earth=R

distance of earth from the sun=D

then mass of planet X=M/2

radius of planet X=4*R

distance of planet X from sun=10*D

part A:

gravitational force between two objects of mass m1 and m2 separated by distance d is given by

F=G*m1*m2/d^2

hence force by sun on earth=F1=G*M*Ms/D^2

where Ms=mass of sun

force by sun on planet X=G*(M/2)*Ms/(10*D)^2

=G*M*Ms/(200*D^2)

=F1/200

hence gravitational force by sun on planet X is weaker by 200 times

part b:

acceleration due to gravity on earth=G*M/R^2=g

acceleration due to gravity on planet X=G*(M/2)/(4*R)^2=G*M/(32*R^2)

=g/32

hence free fall acceleration on planet X near the surface is 1/32 times g.

Q8.

normal force on the block=weight of the block+F

=30*9.8+F

=(294+F) N

maximum static friction force=static friction coefficient*normal force

=0.5*(294+F)

this friction force has to be equal to 200 N atleast so that the block will be at rest.

==>0.5*(294+F)=200

==>294+F=200/0.5=400

==>F=400-294=106 N

hence minimum downward force is 106 N

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