If an object weigh 270 N at the earths surface, what will be the weight of the object at an altitude equal to twice the radius of earth?

Question

If an object weigh 270 N at the earths surface, what will be the weight of the object at an altitude equal to twice the radius of earth? (A) 270 N (B) 90 N (C) 30 N (D) 60 N

Tardigrade Admin · Accepted Answer

Let m is the mass of object and g is acceleration due to gravity at earths surface then mg=270N. The acceleration due to gravity at an altitude of 2Re is, g=(GM/(Re+2Re)2)-=(g/9) So, required weight =mg=(mg/9)=30 N

Q. If an object weigh 270 N at the earths surface, what will be the weight of the object at an altitude equal to twice the radius of earth?

270 N

90 N

30 N

60 N

Let m is the mass of object and g is acceleration due to gravity at earths surface then $m g = 270 N .$ The acceleration due to gravity at an altitude of $2 R_{e}$ is, $g = \frac{GM}{( R _{e} + 2 R _{e} ) ^{2}} - = \frac{g}{9}$ So, required weight $= m g = \frac{m g}{9} = 30 N$

Q. If an object weigh 270 N at the earths surface, what will be the weight of the object at an altitude equal to twice the radius of earth?

270 N

90 N

30 N

60 N

Let m is the mass of object and g is acceleration due to gravity at earths surface then mg=270N. The acceleration due to gravity at an altitude of 2Re​ is, g=(Re​+2Re​)2GM​−=9g​ So, required weight =mg=9mg​=30N

Let m is the mass of object and g is acceleration due to gravity at earths surface then $m g = 270 N .$ The acceleration due to gravity at an altitude of $2 R_{e}$ is, $g = \frac{GM}{( R _{e} + 2 R _{e} ) ^{2}} - = \frac{g}{9}$ So, required weight $= m g = \frac{m g}{9} = 30 N$