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Berapa besar gaya gravitasi antara bumi dan bulan? Jika diketahui massa bumi  kg dan massa bulan  kg, jarak pusat bumi dan bulan  m!

Pertanyaan

Berapa besar gaya gravitasi antara bumi dan bulan? Jika diketahui massa bumi 5 comma 97 cross times 10 to the power of 24 kg dan massa bulan 7 comma 35 cross times 10 to the power of 22 kg, jarak pusat bumi dan bulan 3 comma 84 cross times 10 to the power of 8 m!space 

Pembahasan Soal:

Dengan menggunakan persamaan gaya gravitasi Newton, diperoleh besar gaya gravitasi antara bumi dan bulan:

F subscript g equals fraction numerator G times M subscript B times M subscript b over denominator r squared end fraction F subscript g equals fraction numerator 6 comma 67 cross times 10 to the power of negative 11 end exponent times 5 comma 97 cross times 10 to the power of 24 times 7 comma 35 cross times 10 to the power of 22 over denominator open parentheses 3 comma 84 cross times 10 to the power of 8 close parentheses squared end fraction F subscript g equals 1 comma 98 cross times 10 to the power of 20 space straight N

 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Qohar

Mahasiswa/Alumni Universitas Negeri Semarang

Terakhir diupdate 30 April 2021

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Pertanyaan yang serupa

Diketahui massa planet A adalah 4 kali massa planet B, dan jarak antar pusat planet A ke planet B adalah R. Suatu benda uji bermassa M yang berada pada jarak r dari pusat planet A dan pada garis lurus...

Pembahasan Soal:

Gaya gravitasi yang bekerja pada benda uji adalah nol, maka resultan gaya gravitasinya sama dengan nol:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sum F end cell equals 0 row cell F subscript A minus F subscript B end cell equals 0 row cell F subscript A end cell equals cell F subscript B end cell row cell G fraction numerator M subscript A times M over denominator r subscript A squared end fraction end cell equals cell G fraction numerator M subscript B times M over denominator r subscript B squared end fraction end cell row cell fraction numerator 4 M subscript B over denominator r squared end fraction end cell equals cell M subscript B over open parentheses R minus r close parentheses squared end cell row cell 2 over r end cell equals cell fraction numerator 1 over denominator open parentheses R minus r close parentheses end fraction end cell row cell 2 R minus 2 r end cell equals r row cell 2 R end cell equals cell 3 r end cell row r equals cell 2 over 3 R end cell row r equals cell 0 comma 67 R end cell end table end style

Jadi, jawaban yang tepat adalah C.

Roboguru

Tiga buah benda A, B dan C membentuk segitiga siku-siku seperti gambar berikut! Tentukan besar gaya gravitasi pada benda B!

Pembahasan Soal:

Diketahui:

begin mathsize 14px style m subscript 1 equals m subscript 2 equals m subscript 3 equals 1 space kg R subscript B A end subscript equals 1 space straight m R subscript B C end subscript equals square root of 2 space straight m end style 

Ditanya: FB

Jawab:

Benda B ditarik A menghasilkan FBA dan ditarik benda C menghasilkan FBC dimana sudut yang terbentuk antara FBA dan FBC adalah 60o.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell F subscript B A end subscript end cell equals cell double-struck G fraction numerator m subscript A times m subscript B over denominator R subscript B A end subscript squared end fraction end cell row blank equals cell double-struck G fraction numerator 1 times 1 over denominator 1 squared end fraction end cell row blank equals double-struck G row cell F subscript B C end subscript end cell equals cell double-struck G fraction numerator m subscript C times m subscript B over denominator R subscript B C end subscript squared end fraction end cell row blank equals cell double-struck G fraction numerator 1 times 1 over denominator open parentheses square root of 2 close parentheses squared end fraction end cell row blank equals cell 1 half double-struck G end cell end table end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell F subscript B end cell equals cell square root of F subscript B A end subscript squared plus F subscript B C end subscript squared plus 2 times F subscript B A end subscript times F subscript B c end subscript times cos space theta end root end cell row blank equals cell square root of double-struck G squared plus open parentheses 1 half double-struck G close parentheses squared plus 2 times double-struck G times 1 half double-struck G times cos space 60 end root end cell row blank equals cell square root of double-struck G squared plus 1 fourth double-struck G squared plus 2 times double-struck G times 1 half double-struck G times 1 half end root end cell row blank equals cell square root of double-struck G squared plus 1 fourth double-struck G squared plus 1 half double-struck G squared end root end cell row blank equals cell square root of 7 over 2 double-struck G squared end root end cell row blank equals cell 1 half double-struck G square root of 7 end cell end table end style
    
Jadi, jawaban yang tepat adalah begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 half end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank double-struck G end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 7 end cell end table end style.

Roboguru

Benda A = 16 kg berada pada jarak 9 m dari benda B = 25 kg, sedangkan benda C = 5 kg di antara benda A dan B. Jika gaya gravitasi pada benda C sama dengan nol, maka tentukan jarak antara benda A dan C...

Pembahasan Soal:

Diketahui:

begin mathsize 14px style m subscript A equals 16 space kg m subscript B equals 25 space kg m subscript C equals 5 space kg  r subscript A B end subscript equals 9 space straight m  F subscript A C end subscript equals F subscript C B end subscript G fraction numerator m subscript A m subscript C over denominator r subscript A C end subscript squared end fraction equals G fraction numerator m subscript C m subscript B over denominator open parentheses 9 minus r subscript A C end subscript close parentheses squared end fraction fraction numerator 16 cross times 5 over denominator r subscript A C end subscript squared end fraction equals fraction numerator 5 cross times 25 over denominator open parentheses 9 minus r subscript A C end subscript close parentheses squared end fraction fraction numerator 4 square root of 5 over denominator r subscript A C end subscript end fraction equals fraction numerator 5 square root of 5 over denominator 9 minus r subscript A C end subscript end fraction 4 over r subscript A C end subscript equals fraction numerator 5 over denominator 9 minus r subscript A C end subscript end fraction 4 open parentheses 9 minus r subscript A C end subscript close parentheses equals 5 r subscript A C end subscript 36 equals 9 r subscript A C end subscript bold italic r subscript bold A bold C end subscript bold equals bold 4 bold space bold m end style  

Jadi, jawaban yang tepat adalah 4 mundefined 

Roboguru

Benda P yang bermassa 1 kg berada pada jarak 6 meter dari benda R yang bermassa 4 kg. Benda S bermassa 2 kg berada di antara benda P dan R. Jika gaya gravitasi yang dirasakan benda S sama dengan nol, ...

Pembahasan Soal:

Diketahui:

begin mathsize 14px style m subscript P equals 1 space kg m subscript R equals 4 space kg m subscript S equals 2 space kg r subscript P R end subscript equals 6 space straight m  F subscript P S end subscript equals F subscript S R end subscript G fraction numerator m subscript P m subscript S over denominator r subscript P S end subscript squared end fraction equals G fraction numerator m subscript S m subscript R over denominator left parenthesis 6 minus r subscript P S end subscript right parenthesis squared end fraction m subscript P over r subscript P S end subscript squared equals m subscript R over left parenthesis 6 minus r subscript P S end subscript right parenthesis squared fraction numerator square root of m subscript P end root over denominator r subscript P S end subscript end fraction equals fraction numerator square root of m subscript R end root over denominator 6 minus r subscript P S end subscript end fraction r subscript P S end subscript equals fraction numerator 6 square root of m subscript P end root over denominator square root of m subscript P end root plus square root of m subscript R end root end fraction r subscript P S end subscript equals fraction numerator 6 square root of 1 over denominator square root of 1 plus square root of 4 end fraction bold italic r subscript bold italic P bold italic S end subscript bold equals bold 2 bold space bold m end style 

Jadi, jarak antara P dan S adalah 2 mspace 

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Suatu segitiga sama sisi mempunyai sisi 5 meter dan setiap sudutnya diberi benda bermassa 10 kg, 20 kg, dan 30 kg. Apabila G = konstanta gravitasi = , benda yang mempunyai massa 10 kg mengalami gaya s...

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell R subscript 12 end cell equals cell 5 space straight m end cell row cell straight R subscript 13 end cell equals cell 5 space straight m end cell row cell m subscript 1 end cell equals cell 10 space kg end cell row cell straight m subscript 2 end cell equals cell 20 space kg end cell row cell straight m subscript 3 end cell equals cell 30 space kg end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell F subscript 12 end cell equals cell fraction numerator G times m subscript 1 times m subscript 2 over denominator R subscript 12 squared end fraction end cell row cell F subscript 12 end cell equals cell fraction numerator G times 10 times 20 over denominator 5 squared end fraction end cell row cell F subscript 12 end cell equals cell 8 space G end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell F subscript 13 end cell equals cell fraction numerator G times m subscript 1 times m subscript 3 over denominator R subscript 13 squared end fraction end cell row cell F subscript 13 end cell equals cell fraction numerator G times 10 times 30 over denominator 5 squared end fraction end cell row cell F subscript 13 end cell equals cell 12 space G end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell F subscript r end cell equals cell square root of F subscript 12 squared plus F subscript 13 squared plus 2 times F subscript 12 times F subscript 13 times cos space theta end root end cell row cell F subscript r end cell equals cell square root of left parenthesis 8 space G right parenthesis squared plus left parenthesis 12 space G right parenthesis squared plus 2 times left parenthesis 8 space G right parenthesis times left parenthesis 12 space G right parenthesis times cos space 60 end root end cell row cell F subscript r end cell equals cell square root of 64 space G squared plus 144 space G squared plus 192 space G squared times 1 half end root end cell row cell F subscript r end cell equals cell square root of 208 space G squared plus 96 space G squared end root end cell row cell F subscript r end cell equals cell square root of 304 space G squared end root end cell row cell F subscript r end cell equals cell 17 comma 4 space G end cell row cell F subscript r end cell equals cell 17 comma 4 cross times 6 comma 673 cross times 10 to the power of negative 11 end exponent end cell row cell F subscript r end cell equals cell 116 comma 1 cross times 10 to the power of negative 11 end exponent space straight N end cell row cell F subscript r end cell almost equal to cell 1 comma 2 cross times 10 to the power of negative 9 end exponent space straight N end cell end table

Jadi, jawaban yang tepat adalah E

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