Question

Q.2
Match the pair
1
i. Uniform motion
N
Beyond gravitational limit
ii. Escape velocity
a=0 m/s2
Radial towards centre
Joule
Increases with height
m/s
Object at rest​

Answer 1

Answer:

hello dear/

Explanation:

The g at poles is given by g  

p

​  

=  

R  

2

 

GM

​  

 

Now, g at equator, g  

e

​  

=  

2

g  

p

​  

 

​  

=  

2R  

2

 

GM

​  

 

But, acceleration on body at equator =g  

p

​  

−  

R

V  

2

 

​  

⇒  

R  

2

 

GM

​  

−  

R

V  

2

 

​  

=  

2R  

2

 

GM

​  

 

solving, we get V=(  

2R  

2

 

GM

​  

)  

0.5

 

Now, escape velocity at poles, V  

e

​  

=(  

R  

2

 

2GM

​  

)  

0.5

=(4×  

R  

2

 

GM

​  

)  

0.5

=2V.