Please, I need help with this problem. I would really appreciate it.
The top of Mt. Everest is 8850 m above sea level. Assume that sea level is at the average Earth radius of 6.38×106 m. What is the magnitude of the gravitational acceleration at the top of Mt. Everest? The mass of the Earth is 5.97×1024 kg.
Answer:__________ m/s^2
Thank you.
3 answers
I just did this for someone a few minutes ago.
Posted by Kathy on Sunday, April 21, 2013 at 6:37pm.
Could anyone please help me with this problem? I would really appreciate it.
The top of Mt. Everest is 8850 m above sea level. Assume that sea level is at the average Earth radius of 6.38×106 m. What is the magnitude of the gravitational acceleration at the top of Mt. Everest? The mass of the Earth is 5.97×1024 kg.
Answer ________ N
Thank you.
physics (Newton - Damon, Sunday, April 21, 2013 at 6:52pm
F = G m M / r^2
a = F/m
a = G M/r^2
a = 6.67*10^-11 (5.97*10^24) / (6,380,000+8,850)^2
a = (6.67*5.97/4.08)10^(-11+24-13)
a = 9.76 * 10^0
= 9.76 m/s^2
Could anyone please help me with this problem? I would really appreciate it.
The top of Mt. Everest is 8850 m above sea level. Assume that sea level is at the average Earth radius of 6.38×106 m. What is the magnitude of the gravitational acceleration at the top of Mt. Everest? The mass of the Earth is 5.97×1024 kg.
Answer ________ N
Thank you.
physics (Newton - Damon, Sunday, April 21, 2013 at 6:52pm
F = G m M / r^2
a = F/m
a = G M/r^2
a = 6.67*10^-11 (5.97*10^24) / (6,380,000+8,850)^2
a = (6.67*5.97/4.08)10^(-11+24-13)
a = 9.76 * 10^0
= 9.76 m/s^2
Hey, look for answers already posted.
1 answer